There are n electrons of charge e on a drop of oil of density $\rho .$ It is in equilibrium in an electric field E. Then radius of drop is

Given data: n electrons e on a drop of oil of density $\rho .$

Concept used: electric field and electric fields lines 

Detailed solution:

Due to the cancelling out of the gravity force and the electric operating on it, the drop is in equilibrium. Electrical power = $E\times ne$

Gravitational force = 

$$\begin{gathered} \frac{4}{3}\pi r^3\rho g \\ =\frac{4}{3}\pi r^3\rho g=\text{ Ene }\Rightarrow r=\sqrt[3]{\frac{3Ene}{4\pi \rho g}} \end{gathered}$$