We know that on any two points of equipotential surface, potential different is zero or of equal potential.
$$\therefore E=\dfrac{-d V}{d r}$$
So the electric field intensity is inversely proportional to the separation between equipotential surfaces.
So, equipotential surfaces are closer in regions of large electric fields. Thus, it
verifies answer(a)
The electric field is larger near the sharp edge, due to larger charge density as A is very small
$$\because\sigma=\dfrac{q}{A}$$
So equipotential surfaces are closer or crowded. It verifies answer (b).
As the electric field $$E=\dfrac{k q}{r^{2}}$$ and potential or field decreases as size of body increases or vice- versa (case of earth), so the equipotential
surfaces will be more crowded if the charge density $$\sigma=\dfrac{q}{A}$$
increases. It verifies the answer (c).
As the equipotential surface depends on distance $$r$$ by $$E=\dfrac{-d V}{r}$$ and $$V=\dfrac{k q}{r}$$
Equipotential surface depends on charge density at that place which is different at different place, so equipotential surface are not equispaced all over.