At a distance of 5 cm and 10cm outwards from the surface of a uniformly charged solid sphere, the potentials are 100V and 75V respectively then

Potential at 5cm from the surface = 100V 
Potential at 10cm from surface = 75V
$\begin{array}{l}\frac{\mathrm{Q}\left(100\right)}{4{\mathrm{\pi ϵ}}_0(\mathrm{x}+5)}=100\\ \frac{\mathrm{Q}\left(100\right)}{4{\mathrm{\pi ϵ}}_0(\mathrm{x}+10)}=75\\ \Rightarrow \frac{\mathrm{x}+10}{\mathrm{x}+5}=\frac{4}{3}\Rightarrow 3\mathrm{x}+30=4\mathrm{x}+20\\ \Rightarrow \mathrm{x}=10\mathrm{cm}\\ \therefore \frac{\mathrm{Q}\times 9\times {10}^9\times {10}^2}{15}=100\\ \mathrm{Q}=\frac{15}{9}\times {10}^{-9}=\frac{5}{3}\times {10}^{-9}\mathrm{C}\end{array}$
Potential at surface $=\frac{\mathrm{Q}\times 100}{4{\mathrm{\pi ϵ}}_0\times 10}$
$\begin{array}{l}=\frac{9\times {10}^9\times \frac{5}{3}\times {10}^{-9}\times 100}{10}\\ =150\mathrm{V}\end{array}$
Potential at centre $\frac{3}{2}{\mathrm{V}}_{\mathrm{s}}$
$=\frac{3}{2}\times 150=225\mathrm{V}$
Field on surface $=\frac{150}{10\times {10}^{-2}}$
$=1500 \mathrm{V}/\mathrm{m}$