You question isn't specific enough; it needs a little work to clarify the fusion setup. For example, what fuel type are you talking about fusing? Is there confinement, so that this is a thermal fusion reaction, or would just one fusion reaction be sufficient?
For example, the temperature required to overcome the Coulomb barrier for deuterium-tritium fusion is $4.5 \times 10^7 \mathrm{K}$, which corresponds to an energy per particle of $E \approx\frac{3}{2} k_\mathrm{B} T$ $\approx 10^{-15} \mathrm{J}$. So the simplest way to answer your question, then, is to say that you just need to give this much energy to a deuteron, and direct it perfectly towards its target tritium. And you could do that in principle with only that much solar energy. But the area needed for that would depend on how long you apply the force to the deuteron, because the sun is giving you power (which is energy per unit time).
Now, that's probably cheating -- though I can't say for sure, because your question isn't specific enough. Let's say you actually want to just heat up some sample that is magically contained, but still lets all that solar energy in. Basically, you need to balance the heat output against the heat input. Modeling the magical ball of fuel as a blackbody of area $a$, you can use the Stefan-Boltzman law to say that the power output just from blackbody radiation is
$$
I_\mathrm{out} = a\, \sigma\, T^4.
$$
If you collect solar energy over an area $A$, and perfectly transfer all of that energy to your fuel, the energy input is in the neighborhood of
$$
I_\mathrm{in} = A\, (1 \mathrm{kW}/\mathrm{m}^2).
$$
You need to balance the output and input energies to get the temperature $T$, so
$$
T = \sqrt[4]{\frac{A}{a\, \sigma}\, (1 \mathrm{kW}/\mathrm{m}^2)} \geq 4.5 \times 10^7 \mathrm{K}.
$$
So you can see that the area $A$ over which you have to collect energy is inversely proportional to the area $a$ of your ball (or torus) of fuel, according to this equation. For example, if your fuel is a cube of side $1\mathrm{cm}$, it has area $8 \times 10^{-4}\mathrm{m}^2$
$$
A \gtrsim (4\times 10^{30}\mathrm{K}^4) (8 \times 10^{-4}\mathrm{m}^2) (6 \times 10^{-8}\, \mathrm{\mathrm{W}\, \mathrm{m}^{-2}\mathrm{K}^{-4}}) / (10^3 \mathrm{W}/\mathrm{m}^2) \approx 10^{17} \mathrm{m}^{2}
$$
That's a good deal larger than the earth. In fact, it's larger than the surface area of the entire sphere surrounding the sun at the radius of Earth's orbit. So this says that the sun doesn't provide enough energy to ignite a blackbody in the shape of a cube of side $1\mathrm{cm}$. This suggests to me that (1) my math is wrong; (2) you'd have to have a smaller pellet of fuel; (3) you'd have to magically recycle that blackbody radiation somehow; or (4) you'd need a different star.
But of course, all of this is still ignoring what may be your motivation for asking this question. This presumably would be a very poor method for capturing energy, because the energy released by fusion would need to be captured. And any way I know of doing this would be easier to do just by directly converting the solar energy.
