Answer
$t = 6.9\times 10^{-3}~s$
Work Step by Step
We can find the time when the charge is $10~\mu C$:
$Q(t) = Q_0~e^{-t/RC}$
$\frac{Q(t)}{Q_0} = e^{-t/RC}$
$ln(\frac{Q(t)}{Q_0}) = -t/RC$
$t = -RC\times ln(\frac{Q(t)}{Q_0})$
$t = -(1.0\times 10^3~\Omega)(10\times 10^{-6}~F)\times ln(\frac{10~\mu C}{20~\mu C})$
$t = 6.9\times 10^{-3}~s$
