1. Potential on the surface of a uniform solid sphere

For a solid sphere of mass $M$ and radius $R$, the gravitational potential at the surface is

$V_s=-\frac{GM}{R}$

The question says this surface potential is $V$. So,

$V=-\frac{GM}{R}$

2. Potential at the center of a uniform solid sphere

For a uniform solid sphere, the gravitational potential at a distance $r$ from the center is

$V(r)=-\frac{GM}{2R^3}(3R^2-r^2)$

At the center, $r=0$:

$V_c=-\frac{GM}{2R^3}(3R^2)= -\frac{3GM}{2R}$

3. Relate center potential to surface potential

Since

$V=-\frac{GM}{R}$

therefore

$V_c=\frac{3}{2}\left(-\frac{GM}{R}\right)=\frac{3V}{2}$

4. Choose the correct option

Thus, the gravitational potential at the center is

$\boxed{\frac{3V}{2}}$

So the correct option is A.

Note: Gravitational potential is negative. Hence if surface potential is $V$, center potential is also more negative, and algebraically equal to $\frac{3V}{2}$.