#### Answer

The diameter of each disk is 18.6 mm

#### Work Step by Step

We can find magnitude of the charge that was transferred.
$Q = (3.0\times 10^9)(1.6\times 10^{-19}~C)$
$Q = 4.8\times 10^{-10}~C$
We can find the area of each disk.
$E = \frac{Q}{A~\epsilon_0}$
$A = \frac{Q}{E~\epsilon_0}$
$A = \frac{4.8\times 10^{-10}~C}{(2.0\times 10^5~N/C)(8.85\times 10^{-12}~C^2/N~m^2)}$
$A = 2.71\times 10^{-4}~m^2$
We can find the radius of each disk.
$\pi~r^2 = A$
$r = \sqrt{\frac{A}{\pi}}$
$r = \sqrt{\frac{2.71\times 10^{-4}~m^2}{\pi}}$
$r = 0.0093~m = 9.3~mm$
Since the diameter is twice the radius, the diameter of each disk is 18.6 mm