b
The electric potential is a scalar quantity. So the potential at the center is the sum of potential due to $(n-1) q$ number of charges.i.e, $V=k \frac{(n-1) q}{r}$
The electric filed is a vector quantity. So the electric field cancel each other for the charges of opposite corner of polygon. Only charge $n q-(n-1) q=q$ will contribute
the electric field at the center of polygon. thus, $E=k \frac{q}{r^{2}}$
$\therefore \frac{V}{E}=k \frac{(n-1) q}{r} \times \frac{r^{2}}{k q}=r(n-1)$