A point charge Q is located on the axis of disc of radius R at a distance b from the plane of the disk. If one fourth of the electric flux from the charge passes through the disk, the radius of disk is

Total flux associated with charge Q for solid angle $4\mathrm{\pi } \mathrm{is} \frac{\mathrm{Q}}{{\mathrm{\epsilon }}_0}$.

Solid angle subtended by disc at Q, $\mathrm{\Omega }=2\mathrm{\pi }(1-\mathrm{cos\alpha })=2\mathrm{\pi }\left[1-\frac{\mathrm{b}}{\sqrt{{\mathrm{b}}^2+{\mathrm{R}}^2}}\right]$

$$\begin{gathered} \text{ Flux through disc }=\frac{\mathrm{Q\Omega }}{{\mathrm{\epsilon }}_{0}4\mathrm{\pi }} \\ \text{ It is given to be }\frac{\mathrm{Q}}{4{\mathrm{\epsilon }}_0} \\ \text{ So }\frac{\mathrm{Q}}{4{\mathrm{\epsilon }}_0}=\frac{\mathrm{Q\Omega }}{{\mathrm{\epsilon }}_{0}4\mathrm{\pi }} \\ \text{ Solve above, we get }\mathrm{R}=\sqrt{3}\mathrm{b} \end{gathered}$$