In a hollow spherical shell, potential (V) changes with respect to distance (s) from centre as

In shell, q charge is uniformly distributed over its surface, it behaves as a conductor.

V = potential at surface $=\frac{q}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{R}}$ and inside $V=\frac{q}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{R}}$

Because of this it behaves as an equipotential surface.

Outside the shell $V=\frac{q}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{r}}, r is the dis\tan ce from the centre.$

Option 2 will best representation.