Answer
$C = 18~\mu F$
Work Step by Step
We can find the value $C$ of the capacitor:
$I(t) = I_0~e^{-t/RC}$
$0.25~I_0 = I_0~e^{-t/RC}$
$0.25 = e^{-t/RC}$
$ln(0.25) = -\frac{t}{RC}$
$C = -\frac{t}{R~ln(0.25)}$
$C = -\frac{2.5\times 10^{-3}~s}{(100~\Omega)\times ln(0.25)}$
$C = 1.8\times 10^{-5}~F$
$C = 18~\mu F$
