Given
E: Electric field
E = 402 N/C
m: mass
m = 3.28 x 10-17 kg
Procedure
Let q be the amount of the charge that the drop carries.
The coulomb force balances the gravitational force acting on the drop at equilibrium.
[tex]\begin{gathered} qE=mg \\ q=\frac{mg}{E} \\ q=\frac{3.28\times10^{-17}kg*9.8m/s^2}{402\text{ N/C}} \\ q=8.09\times10^{-19}C \end{gathered}[/tex]The answer would be 8.09x10^-19 C