Answer
(a) $q=52.0 \times 10^{-9} \mathrm{C}$
(b) $V = 3.75 \,\text{kV}$
Work Step by Step
(a) The charge $q$ on the sphere could be calculated by knowing the value of the potential
\begin{gather}
V =\frac{1}{4 \pi \epsilon_{0}} \frac{q}{R} \\
q =4 \pi \epsilon_{0} R V\\
q =\left(4\pi \times 8.85 \times 10^{-12} \mathrm{F/m} \right)(0.125 \mathrm{m})\left(3.75 \times 10^{3} \mathrm{V}\right)\\
\boxed{q=52.0 \times 10^{-9} \mathrm{C}}
\end{gather}
(b) The potential on the surface is the same for the potential for the center of the sphere
$$V = 3.75 \,\text{kV}$$
