If on the concentric hollow spheres of radii $r$ and $R(>r)$ the charge $Q$ is distributed such that their surface densities are same then the potential at their common centre is

Let the charges on the two spheres be $q_1$ and $q_2$. Since the surface charge densities are equal, $$ \frac{q_1}{4\pi r^2}=\frac{q_2}{4\pi R^2} $$ So, $$ \frac{q_1}{q_2}=\frac{r^2}{R^2} $$ Also the total charge is $Q$, hence $$ q_1+q_2=Q $$ Using the ratio, $$ q_1=\frac{Qr^2}{r^2+R^2} $$ $$ q_2=\frac{QR^2}{r^2+R^2} $$ Potential at the centre due to a spherical shell is charge divided by $4\pi\varepsilon_0$ times its radius. Therefore, $$ V=\frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r}+\frac{q_2}{R}\right) $$ Substituting, $$ V=\frac{1}{4\pi\varepsilon_0}\left(\frac{Qr^2}{r(r^2+R^2)}+\frac{QR^2}{R(r^2+R^2)}\right) $$ $$ V=\frac{1}{4\pi\varepsilon_0}\cdot \frac{Q(r+R)}{r^2+R^2} $$ This matches option $\frac{Q(R+r)}{4\pi\varepsilon_0(R^2+r^2)}$ after re-checking all options.