If you consider first time dilation due to relative velocity in flat spacetime, then relative velocity always makes the moving clock run slower. If you're interested, I explain why this in the first part of my answer to Is gravitational time dilation different from other forms of time dilation?. So in flat spacetime in $A$'s rest frame there is no way for $B$ to move that will make $B$'s clock run faster.
But the situation is different when the time dilation is due to a gravitational field. To calculate the time dilation exactly you need to know the metric, but there is a useful approximation that links the time dilation to the gravitational potential (per unit mass):
$$ \frac{dt_B}{dt_A} = \sqrt{1 - \frac{2(\Phi_A - \Phi_B)}{c^2}} \tag{1} $$
The quantity $\Phi_A - \Phi_B$ is the difference in gravitational potential energy between $A$ and $B$, and $dt_B/dt_A$ is the time dilation of $B$'s clock relative to $A$'s clock i.e. if this is less than one $B$'s clock runs slower and if it's greater than one $B$'s clock runs faster.
To make this concrete let's take $B$ to be hovering 1m above $A$. You may have come across the expression for the change in potential energy in a uniform gravitational field with acceleration $g$:
$$ \Delta\Phi = mgh $$
In this case $\Phi_B < \Phi_A$, i.e. it is less negative, and we need to divide by $m$ to get the energy per unit mass, so putting in $h = 1$m we get:
$$ \Phi_A - \Phi_B = -g $$
and substituting in equation (1) and using Earth's gravitational acceleration, $g = 9.8$m/s$^2$, gives:
$$\begin{align}
\frac{dt_B}{dt_A} &= \sqrt{1 + \frac{2g}{c^2}} \\
&\approx 1.0000000000000001
\end{align}$$
This is very small but it is greater than one i.e. $B$'s clock runs faster than $A$'s clock.
However we can't make $B$'s clock run arbitrarily fast. The best we can do is take $B$ to well away from the Earth so $\Phi_B \approx 0$. If $A$ is on the Earth's then $\Phi_A$ is given by the Newtonian expression:
$$ \Phi_A = -\frac{GM_E}{r_E} $$
where $M_E$ is the mass of the Earth and $r_E$ is the radius of the Earth. This gives $\Phi_A \approx 6.25 \times 10^7$, and putting this into equation (1) we get:
$$ \frac{dt_B}{dt_A} \approx 1.000000001 $$
which is still pretty small, but it's large enough that the clocks in GPS satellites run measurably faster than clocks on the Earth's surface.
To make the time difference any greater you'd have to move $A$'s clock to somewhere with a lower (i.e. more negative) gravitational potential. For example you could move $A$ to near the event horizon of a black hole.
