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

Given potential at 5cm from surface outer = 100V.

Potential at 10 cm from surface = 75 V

Let the radius of sphere be 'x',

$\frac{\mathrm{Q}\left(100\right)}{4{\mathrm{\pi ϵ}}_0(\mathrm{x}+5)}=100,\frac{\mathrm{Q}\left(100\right)}{4{\mathrm{\pi ϵ}}_0(\mathrm{x}+10)}=75$

 $$\begin{gathered} \Rightarrow \frac{x+10}{x+5}=\frac{4}{3}, \\ \Rightarrow 3x+30=4x+20 \end{gathered}$$

$\Rightarrow \mathrm{x}=10 \mathrm{cm}$

On substituting, $\frac{\mathrm{Q}\times 9\times {10}^9\times {10}^2}{\left(15\right)}=100$

$\Rightarrow \mathrm{Q}=\frac{15}{9}\times {10}^{-9}=\frac{5}{3}\times {10}^{-9} \mathrm{C}$

Potential at surface ${\mathrm{V}}_{\mathrm{S}}=\frac{\mathrm{Q}\times 100}{4{\mathrm{\pi ϵ}}_0\left(10\right)}=150 \mathrm{V}$ and 

Potential at centre $=\frac{3}{2}{\mathrm{V}}_{\mathrm{S}}=225 \mathrm{V}$

field on surface $=\frac{150}{10\times {10}^{-2}}=1500 \mathrm{V}/\mathrm{m}$