A cube of side $x$ has charge $q$ at each of its vertices. The electric potential due to these charges at the center of the cube is

The length of diagonal of the cube of each side $x$ is $\sqrt{3x^2}=x\sqrt{3}$

$\therefore$ Distance between centre of the cube and each vertex, $r=\frac{x\sqrt{3}}{2}$

Now, potential, $V=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{r}$

Since cube has $8$ vertices and $8$ charges each of value $q$ are present there

$\therefore V=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{8q}{{\displaystyle \frac{x\sqrt{3}}{2}}}=\frac{4q}{\sqrt{3}{\mathrm{\pi \epsilon }}_0\mathrm{x}}$