c
We know, \(\mathrm{V}_{0}=\frac{\mathrm{Kq}}{\mathrm{R}}=\mathrm{V\,surface}\)

Now, \(\mathrm{V}_{\mathrm{i}}=\frac{\mathrm{Kq}}{2 \mathrm{R}^{3}}\left(3 \mathrm{R}^{2}-\mathrm{r}^{2}\right) \quad[\mathrm{For}\, \mathrm{r}<\mathrm{R}]\)

At the centre of sphare \(r=0 .\)

Here \(\mathrm{V}=\frac{3}{2} \mathrm{V}_{0}\)

Now, \(\frac{5}{4} \frac{\mathrm{Kq}}{\mathrm{R}}=\frac{\mathrm{Kq}}{2 \mathrm{R}^{3}}\left(3 \mathrm{R}^{2}-\mathrm{r}^{2}\right)\)

\(\mathrm{R}_{2}=\frac{\mathrm{R}}{\sqrt{2}}\)

\(\frac{3}{4} \frac{\mathrm{Kq}}{\mathrm{R}}=\frac{\mathrm{Kq}}{\mathrm{R}^{3}}\)

\(\frac{1}{4} \frac{\mathrm{Kq}}{\mathrm{R}}=\frac{\mathrm{Kq}}{\mathrm{R}_{4}}\)

\(\mathrm{R}_{4}=4 \mathrm{R}\)

Also, \(\mathrm{R}_{1}=0\) and \(\mathrm{R}_{2}<\left(\mathrm{R}_{4}-\mathrm{R}_{3}\right)\)