Answer
$V = 20~V$
Work Step by Step
We can find the stored energy after $2.0~\mu s$:
$E = P~t = (200)(2.0\times 10^{-6}~s) = 4.0\times 10^{-4}~J$
We can find the voltage:
$U = \frac{1}{2}CV^2$
$V^2 = \frac{2U}{C}$
$V = \sqrt{\frac{2U}{C}}$
$V = \sqrt{\frac{(2)(4.0\times 10^{-4}~J)}{2.0\times 10^{-6}~F}}$
$V = 20~V$
