Answer
The charge on the drop is $~~-5.0~e$
Work Step by Step
We can find the mass of the drop:
$M = \rho~V$
$M = \frac{4}{3}\pi R^3~\rho$
$M = (\frac{4}{3})(\pi) (1.64\times 10^{-6}~m)^3~(851~kg/m^3)$
$M = 1.5724\times 10^{-14}~kg$
If the drop is suspended, then the upward force from the electric field must be equal in magnitude to the drop's weight.
We can find the magnitude of the charge on the drop:
$q~E = Mg$
$q = \frac{Mg}{E}$
$q = \frac{(1.5724\times 10^{-14}~kg)(9.8~m/s^2)}{1.92\times 10^5~N/C}$
$q = 8.0258\times 10^{-19}~C$
$q = (8.0258\times 10^{-19}~C)(\frac{1~e}{1.6\times 10^{-19}~C})$
$q = 5.0~e$
Since the force from the electric field is directed upward, the charge on the drop must be negative.
The charge on the drop is $~~-5.0~e$
