${\rm{A}} = \frac{1}{{4\pi { \in _0}}} \cdot \frac{{\rm{Q}}}{{\rm{R}}};$ The potential of $\mathrm{q}$ at the surface
$A = \frac{1}{{4\pi { \in _0}}} \cdot \frac{q}{R}$
The potential at $\mathrm{B}$ is due to $\mathrm{Q}$ inside $ = \frac{1}{{4\pi { \in _0}}} \cdot \frac{{\rm{Q}}}{{\rm{R}}}$
The potential at $\mathrm{B}$ due to ${\rm{q}} = \frac{1}{{4\pi { \in _0}}} \cdot \frac{{\rm{q}}}{{\rm{r}}}$
$\therefore $ Potential at ${\rm{A}},{{\rm{V}}_{\rm{A}}} = \frac{1}{{4\pi { \in _0}}}\left( {\frac{{\rm{Q}}}{{\rm{R}}} + \frac{{\rm{q}}}{{\rm{R}}}} \right)$
Potential at ${\rm{B}},{{\rm{V}}_{\rm{B}}} = \frac{1}{{4\pi { \in _0}}}\left( {\frac{{\rm{Q}}}{{\rm{R}}} + \frac{{\rm{q}}}{{\rm{r}}}} \right)$
$\therefore {{\rm{V}}_{\rm{B}}} - {{\rm{V}}_{\rm{A}}} = \frac{1}{{4\pi { \in _0}}}\left( {\frac{{\rm{q}}}{{\rm{r}}} - \frac{{\rm{q}}}{{\rm{R}}}} \right)$


Reason : The electric field between the plates of a charged isolated capacitance increases when dielectric fills whole space between plates.