As given the surface charge density for both the spheres is same, hence

$$\begin{gathered} {\sigma }_1={\sigma }_2 \\ \therefore \frac{Q_1}{A_1}=\frac{Q_2}{A_2}\Rightarrow \frac{Q_1}{4\pi R^2}=\frac{Q_2}{4\mathrm{\pi }{\left(4\mathrm{R}\right)}^2} \\ \therefore {\mathrm{Q}}_2=16{\mathrm{Q}}_1 \end{gathered}$$

From above diagram, the electric potential for both the spheres is 

${\mathrm{V}}_{\mathrm{Inner}}=\frac{{\mathrm{KQ}}_1}{\mathrm{R}}+\frac{{\mathrm{KQ}}_2}{4\mathrm{R}}=\frac{\mathrm{K}}{\mathrm{R}}\left({\mathrm{Q}}_1+\frac{16{\mathrm{Q}}_1}{4}\right)=\frac{5{\mathrm{KQ}}_1}{\mathrm{R}}.$

In the same way for outer sphere, 

${\mathrm{V}}_{\mathrm{Outer}}=\frac{{\mathrm{KQ}}_1}{4\mathrm{R}}+\frac{{\mathrm{KQ}}_2}{4\mathrm{R}}=\frac{\mathrm{K}}{4\mathrm{R}}\left(16{\mathrm{Q}}_1+{\mathrm{Q}}_1\right)=\frac{17{\mathrm{KQ}}_1}{4\mathrm{R}}$.

Now the potential difference is,

$∆\mathrm{V}={\mathrm{V}}_{\mathrm{inner}}-{\mathrm{V}}_{\mathrm{outer}}$

$=\frac{5{\mathrm{KQ}}_1}{\mathrm{R}}-\frac{17{\mathrm{KQ}}_1}{4\mathrm{R}}=\frac{3{\mathrm{Q}}_1}{16{\mathrm{\pi \epsilon }}_0\mathrm{R}}$.