Answer
The magnitude of the charge on the drop is $1.6\times 10^{-19}~C$
Work Step by Step
If the drop remains stationary, then the electric force on the drop directed upward must be equal in magnitude to the weight of the drop.
We can find the magnitude of the charge on the drop:
$q~E = mg$
$q~(\frac{V}{d}) = mg$
$q = \frac{mg~d}{V}$
$q = \frac{(1.0\times 10^{-15}~kg)(9.80~m/s^2)(0.16~m)}{9760~J/C}$
$q = 1.6\times 10^{-19}~C$
The magnitude of the charge on the drop is $1.6\times 10^{-19}~C$
