Answer
$q = 2 e$
Work Step by Step
We can find the magnitude of the charge on the drop:
$F = q~E$
$F = q~(\frac{V}{d})$
$q = \frac{F~d}{V}$
$q = \frac{(9.6\times 10^{-16}~N)(0.16~m)}{480~J/C}$
$q = 3.2\times 10^{-19}~C$
We can express the magnitude of the charge in terms of $e$:
$q = \frac{3.2\times 10^{-19}~C}{1.6\times 10^{-19}~C} = 2 e$
