Answer
$\sigma = -9.55\times 10^{-15}~C/m^2$
Work Step by Step
We can find the area of the disk:
$A = 4\pi~R^2$
$A = (4\pi)~(0.020~m)^2$
$A = 0.005027~m^2$
We can find the surface charge density:
$\sigma = \frac{q}{A}$
$\sigma = \frac{-300e}{A}$
$\sigma = \frac{(-300)(1.6\times 10^{-19}~C)}{0.005027~m^2}$
$\sigma = -9.55\times 10^{-15}~C/m^2$
