Solution

As the statement of the problem doesn't clarify if the sphere is conducting or non-conductive, let's consider both cases.
Case 1) Non-Conductive sphere:
Let's assume that charge is uniformly distributed in the volume. Let electrical potential be:
$V(r) = -\int_{\infty}^{r} \vec{E} \cdot d\vec{r}$ (1)
for the center, let's consider $r<<R$, or $r \rightarrow 0$.
Applying Gauss Law:
For $r<R$,
$\int \vec{E} \cdot d\vec{S} = \frac{q_{enc}}{\varepsilon_0}$
but the enclosed charge into a sphere of radius r is related to the charge distribution per unit of volume,
$\rho = \frac{Q}{\frac{4}{3}\pi R^3} = \frac{q_{enc}}{\frac{4}{3}\pi r^3}$
$q_{enc} = Q\left(\frac{r}{R}\right)^3$
so,
$E(r) = \frac{Q r}{4\pi\varepsilon_0 R^3}\;\;\;\;\forall\;r < R$
and for $r > R$, the enclosed charge is Q, then,
$E(r) = \frac{Q}{4\pi\varepsilon_0 r^2}\;\;\;\;\forall\;r>R$ (2)
This mean that function $E(r)$ has two behaviors, depending on values of $r$. According (1) and assuming $r\rightarrow 0$,
$V(r) = -\left(\int_{\infty}^{R} \vec{E} \cdot d\vec{r} + \int_{R}^{r} \vec{E} \cdot d\vec{r}\right)$ (I)
developing,
$V(r) = \frac{Q}{4\pi\varepsilon_0 R} - \frac{Q}{8\pi\varepsilon_0 R^3}(r^2-R^2)$
as $r$ tends to zero,
$V(0) = \frac{3Q}{8\pi\varepsilon_0 R^3}$ (3)
Case 2) Conducting sphere
In this case, charge stays on sphere surface, so the electric field inside of it is null ($E(r) = 0$), so potential is constant inside the sphere and coincides with the value of it on the surface. According to (1) and (2),
$V(R) = -\int_{\infty}^{R} \frac{Q}{4\pi\varepsilon_0 r^2}dr$
$V(0) = V(R) = \frac{Q}{4\pi\varepsilon_0 R}$ (4)
Finally, if sphere is non-conductive $V(0)$ depends on the charge distribution, as we saw in (I), but in the second case (maybe the problem refers specifically to a conducting sphere), it **doesn't depend** on it. If the charge is non-uniformly distributed, the potential on the surface **does change** with the local distribution. For example, charge accumulations in certain areas generate angular variations in the potential. However, outside the sphere, at a great distance, the potential depends only on the total charge $Q$, as if it were a point charge at the center.