d
Potential at $5 \mathrm{cm}$ from surface $=\frac{K Q}{R+5}=100$
Potential at $10 \mathrm{cm}$ from surface
$=\frac{K Q}{R+10}=75 \Rightarrow R=10 \mathrm{cm}$
$\therefore$ Potential at surface $=\frac{K Q}{R}=\frac{100 \times 15}{10}=150 \mathrm{V}$
Electric field on surface $=\frac{K Q}{R^{2}}=\frac{100 \times 15 V \times c m}{100 c m^{2}}$
$=1500 \mathrm{V} / \mathrm{m}$