S is a solid neutral conducting sphere. A point charge q of I x 10^-6 C is placed at point A ,C is the center of sphere, and AB is a tangent. BC = 3 m and AB = 4 m.

Distance AC is given by, AC = 5 m

Potential at due to charge q at A is given by

$\mathrm{V}=\frac{\mathrm{kq}}{\mathrm{AC}}=\frac{9\times {10}^9\times 1\times {10}^{-6}}{5}=1.8\times {10}^3=1.8\mathrm{kV}$

Potential at B is given by

${\mathrm{V}}_{\mathrm{B}}={\left({\mathrm{V}}_{\mathrm{B}}\right)}_{\text{due to }\mathrm{q}}+{\left({\mathrm{V}}_{\mathrm{B}}\right)}_{\text{induced }}$

$\left\{{\left({\mathrm{V}}_{\mathrm{B}}\right)}_{\mathrm{i}}=\text{ Potential at }\mathrm{B}\text{ due to induced charge }\right\}$

Thus $1.8\times {10}^3=\frac{\mathrm{kq}}{\mathrm{AB}}+{\left({\mathrm{V}}_{\mathrm{B}}\right)}_{\mathrm{i}}=2.25\times {10}^3+{\left({\mathrm{V}}_{\mathrm{B}}\right)}_{\mathrm{i}}$

${\left({\mathrm{V}}_{\mathrm{B}}\right)}_{\mathrm{i}}=-0.45\mathrm{kV}$

So (1) and (3) are correct.