A solid sphere of radius R has cavity of radius R/2. The solid part has a uniform charge density $\mathrm{\rho }$ and cavity has no charge. If the electric potential at A and magnitude of electric field at point C are $\frac{5{\mathrm{\rho R}}^2}{{\mathrm{x\epsilon }}_0}$ and $\frac{\mathrm{\rho R}}{{\mathrm{y\epsilon }}_0}$, then value of $\frac{\mathrm{x}}{\mathrm{y}}$is 

Without cavity: The potential at A and electric field at C due to the whole solid sphere are given by
${\mathrm{V}}_{\mathrm{A}}^{\text{'}}=\frac{3}{2}\times \left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right)\frac{(4/3){\mathrm{\pi R}}^3\mathrm{\rho }}{\mathrm{R}}=\frac{{\mathrm{\rho R}}^2}{2{\mathrm{\epsilon }}_0}$
${\mathrm{E}}_{\mathrm{C}}^{\text{'}}=\frac{\mathrm{\rho }}{3{\mathrm{\epsilon }}_0}\overrightarrow{\mathrm{AC}}$
The potential at A and electric field at C due to cavity alone are given by 
${\mathrm{V}}_{\mathrm{A}}^{"}=\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right)\frac{(4/3)\mathrm{\pi }{(\mathrm{R}/2)}^3\mathrm{\rho }}{(\mathrm{R}/2)}=\frac{{\mathrm{\rho R}}^2}{12{\mathrm{\epsilon }}_0}$
Net potential ${\mathrm{V}}_{\mathrm{A}}={\mathrm{V}}_{\mathrm{A}}^{\text{'}}-{\mathrm{V}}_{\mathrm{A}}^{\text{'}\text{'}}=\frac{{\mathrm{\rho R}}^2}{2{\in }_{\mathrm{o}}}-\frac{{\mathrm{\rho R}}^2}{12{\in }_0}=\frac{5{\mathrm{\rho R}}^2}{12{\in }_0}$
$$\begin{gathered} \mathrm{x}=12 \\ {\overline{\mathrm{E}}}_{\mathrm{C}}={\overline{\mathrm{E}}}_{{\mathrm{C}}^{\text{'}}}+{\overline{\mathrm{E}}}_{{\mathrm{C}}^{\text{'}\text{'}}} \\ =\frac{\mathrm{\rho }}{3{\in }_0}[\mathrm{AC}-\mathrm{BC}]=\frac{\mathrm{\rho }}{3{\in }_0}\left(\mathrm{AB}\right)=\frac{\mathrm{\rho }}{3{\in }_0}\left(\frac{\mathrm{R}}{2}\right)=\frac{\mathrm{\rho R}}{6{\in }_0} \\ \mathrm{y}=6 \\ \frac{\mathrm{x}}{\mathrm{y}}=\frac{12}{6}=2 \\ {\mathrm{E}}_{\mathrm{C}}^{"}=\frac{\mathrm{\rho }}{3{\mathrm{\epsilon }}_0}\overrightarrow{\mathrm{BC}} \end{gathered}$$