Answer
$\dfrac{1}{4\pi \epsilon_0}\dfrac{Q}{R}$
Work Step by Step
We know that the electric field at the center of this shell is zero.
And we know that the difference in potential between two points is the source of an electric field.
$$E=\dfrac{\Delta V}{d} $$
where $d$ is the separation distance between the two points, the surface of the shell and its center. So, $d=R$
$$E_{\rm center}=\dfrac{\Delta V}{R}=0 $$
Hence,
$$\Delta V=V_{\rm surface}-V_{\rm center}=0\;\rm V\tag 1$$
Recall that the electric potential on the surface of the shell (at $r=R$) is given by
$$V_{\rm surface}=\dfrac{1}{4\pi \epsilon_0}\dfrac{Q}{R}\tag 2$$
Plug into (1),
$$\dfrac{1}{4\pi \epsilon_0}\dfrac{Q}{R}-V_{\rm center}=0$$
Therefore,
$$\boxed{V_{\rm center}=\dfrac{1}{4\pi \epsilon_0}\dfrac{Q}{R}}$$
