b
Maximum potential is at the surface of the sphere
$\frac{1}{2} m u^{2}=v_{c}-v_{s}$
$=\frac{3}{2} \frac{k q^{2}}{R}-\frac{k q^{2}}{R}$
$\frac{1}{2} m u^{2}=\frac{k q^{2}}{2 R}$
$u^{2}=\frac{k q^{2}}{m R}$
$u=\frac{k q}{\sqrt{m R}}=\frac{q}{\sqrt{4 \pi \varepsilon_{0} m R}}$