The capacitance is defined as $C = Q/V $. So in this formula which charge does $Q$ represents,
for example in spherical capacitor with inner earthing (q+q') charge is also present on outside surface so should we consider it. In other case
in a parallel plate capacitor charge in present on 4 surfaces , so which charges are need to be considered while calculating capacitance?
If you connect a battery across a capacitor it will acquire equal and opposite charges on its plates. These charges will reside on the facing surfaces of the plates. It's difficult, is it not, to devise methods to put charges other than equal and opposite ones, on the plates.
Capacitance, $C$, is defined by $C=Q/V$ in which $Q$ is the magnitude of either of the equal and opposite charges on the plates. The equation isn't applicable in the case of charges other than equal and opposite ones.
Suppose that we have two conductors in close proximity (which might be the two plates of a capacitor). The charge in each conductor is stored not only in the mutual capacitance associated with the two conductors, but also in (usually unintended) capacitances between that conductor and any other conductor in the vicinity (or "infinity"). All three capacitances in a real capacitor are depicted in the figure below.
You can visualize the different capacitances by thinking of the E-field lines emerging from a conductor at positive potential (say conductor 1 on the left): if the field line terminates at conductor 2, it (and the corresponding charges it starts or terminates at) are associated with the mutual capacitance $C$. If the field line terminates at a different conductor (or infinity), the charge on conductor 1 is associated with the parasitic capacitance $C_1$. The total charge $Q_1$ on conductor 1 is the sum of the charges stored in $C$ and $C_1$. We can express the charges on the two conductors mathematically as follows: $$ Q_1 = C_1V_1 + C(V_1 - V_2) $$ $$ Q_2 = C_2V_2 + C(V_2 - V_1) $$ where $V_i$ is the potential at conductor $i$. If we solve these for $V_i$, we get $$q = C(V_1 - V_2) = \frac{c_2 Q_1 - c_1 Q_2}{c_1 + c_2 + c_1c_2} $$ $$q_1 = C_1 V_1 = \frac{c_1 [(1 + c_2)Q_1 + Q_2]}{c_1 + c_2 + c_1c_2} $$ $$q_2 = C_2 V_2 = \frac{c_2 [(1 + c_1)Q_2 + Q_1]}{c_1 + c_2 + c_1c_2} $$ where $c_i = C_i/C$.
The answer to your question is: $q$ (well, more precisely, $\pm q$). This is the charge stored in the mutual capacitance between the conductors. In the case of a practical capacitor, $C$ will dominate (meaning $c_i \ll 1$), the charges $q_i$ in the parasitic capacitors will be small, and so the charges on the two plates will normally be approximately equal in magnitude and opposite in sign. Deviations from this are possible, e.g. if both plates are at a high voltage (with the same sign), but $C$ is still the relevant capacitance, and if necessary, the parasitic capacitances are included separately e.g. in circuit analyses. Note that parasitic capacitances are not exclusive to capacitors: any other conductors in your circuit (like wires) will have these. The large mutual capacitance $C$ between its plates (and the charge $q$ associated with it) is what makes a capacitor special and useful.
Finally, let's see how the inclusion of the parasitic capacitances can explain the charges on the spherical and parallel-plate capacitors you show. For the spherical capacitor, we have $c_1 = 0$ (field lines from the inner conductor only terminate at the other conductor). From the equations above, we get $q=Q_1$, $q_1 = 0$ and $q_2 = Q_1 + Q_2$, just like your diagram shows. For the parallel plate capacitor, we can assume $c_1 = c_2 = c$ and the equations reduce to
$$ q = \frac{Q_1 - Q_2}{2 + c} $$ $$ q_1 = \frac{(1 + c)Q_1 + Q_2}{2 + c} $$ $$ q_2 = \frac{(1 + c)Q_2 + Q_1}{2 + c} $$
If $c \ll 1$ (and only then), we get the charges on the inner and outer surfaces that your second image shows.
Let's look at two simple cases first.
Charge one plate with $+Q$ and the other with $-Q$. This is the normal case, the one that occurs when you put a capacitor in a circuit. The total charge is $0$. $C = Q/V$, where $Q$ is the size of the charge on each plate.
If you pull electrons off one plate and crowd them onto the other, the electrons repel each other. This leaves + charges behind on the + plate, which also repel each other. It takes potential energy to force the charges together.
In a capacitor, the two plates are close to each other. The charges on opposite plates attract each other, reducing this potential energy. It is easier to separate charges if you have a capacitor in a circuit. It still takes energy, and this affects the voltages and currents in the circuit. This can be useful. This is the purpose of a capacitor.
Instead, let's put $+Q$ one each plate. You have a total of $2Q$. This charge has to come from outside the circuit somewhere. Crowding extra electrons into a circuit takes energy. They repel each other and separate as much as possible. Crowding into two closely spaced plates takes more energy than this. It isn't realistic. This isn't how circuits are used, and isn't what capacitors are for.
The case in your example is a combination of the two. An extra charge is put on the capacitor and a voltage is applied so that some normal capacitor-like separation of charge occurs. It means there may be a $-$ charge on both plates, but it is bigger on one than the other. Again this is unrealistic.
There is a point too it. You have to separate out the two different ways of distributing charge. Separations like this are a common thing in physics.
One way is charges are equal and the same on both plates. The other is equal and opposite. Given $Q_1$ on plate $1$ and $Q_2$ on plate $2$, you can figure out how much charge is distributed each way.
It is a confusing thing to do for capacitors. It makes more sense for forces and torques. See Toppling of a cylinder on a block. You can see that separating forces like this would tell you how about forces that accelerate an object and spin an object.
So in this formula which charge does $Q$ represents,
To put it simply, it represents the net positive or negative charge on either plate of a charged capacitor.
I think you may not understand how a charged capacitor is configured. Hopefully, the following will help your understanding.
The top diagram below shows an uncharged capacitor. Note that the number of positive and negative charges on each plate are the same. So the net charge on either plate is zero. Note that only the electrons on each plate are mobile, being able to freely move about the positive charge.
The bottom diagram shows the same capacitor after being "charged" by some voltage source, say a battery. Electrons are taken from the right plate by the positive terminal of the battery, leaving the right plate with a net positive charge. Note that there are more positive charges than negative (count them). The negative terminal of the battery deposits an equal amount of electrons on the left plate, leaving the left plate with a net negative charge (not there are more electrons than protons on the left plate. The total number of electrons and protons on both plates taken together is the same as on the uncharged capacitor. They have simply been redistributed by the voltage source.
Since only the electrons on each plate are free to move, the electrons are on the left plate are attracted to the positive charge on the right plate. The electrons on the right plate are repelled and move to the outer surface of the right plate.
So again to answer your question "What does Q in the capacitance formula represent?", it is simply the net positive or negative charge on either plate that results the capacitor being charged by the battery. In the diagram, there are eight more electrons than protons on the left plate and eight more protons than electrons on the right plate. So $Q=8$. Of course, in reality, we are dealing with billions and billions of protons and electrons, so the value of $Q$ is in Coulombs where 1 Coulomb equals the charge of 6.24 x 10$^{16}$ electrons or protons.
Hope this helps.
