Answer
$\sigma = -3.82\times 10^{-14}~C/m^2$
Work Step by Step
We can find the area of the disk:
$A = \pi~R^2$
$A = (\pi)~(0.020~m)^2$
$A = 0.001257~m^2$
We can find the surface charge density:
$\sigma = \frac{q}{A}$
$\sigma = \frac{-300e}{A}$
$\sigma = \frac{(-300)(1.6\times 10^{-19}~C)}{0.001257~m^2}$
$\sigma = -3.82\times 10^{-14}~C/m^2$
