Answer
$r = 5.21\times 10^{-7}~m$
Work Step by Step
The force exerted upward on the droplet by the electric field must be equal in magnitude to the weight of the droplet.
We can find the mass of the droplet:
$mg = E~q$
$mg = \frac{\Delta V~q}{d}$
$m = \frac{\Delta V~q}{g~d}$
$m = \frac{(25~V)(15)(1.6\times 10^{-19}~C)}{(9.8~m/s^2)~(12\times 10^{-3}~m)}$
$m = 5.1\times 10^{-16}~kg$
We can find the radius of the droplet:
$\rho = \frac{m}{V}$
$V = \frac{m}{\rho}$
$\frac{4}{3}~\pi~r^3 = \frac{m}{\rho}$
$r^3 = \frac{3m}{4\pi~\rho}$
$r = \sqrt[3] {\frac{3m}{4\pi~\rho}}$
$r = \sqrt[3] {\frac{(3)(5.1\times 10^{-16}~kg)}{(4\pi)(860~kg/m^3)}}$
$r = 5.21\times 10^{-7}~m$
