Answer
$145.3\times 10^{-6}C$
Work Step by Step
We can find the required amount of charge as follows:
$q(t)=C\mathcal{E}(1-e^{\frac{-t}{RC}})$
We plug in the known values to obtain:
$q(4.2\times 10^{-3}s)=(23\times 10^{-6}F)(9.0V)[1-e^{\frac{-4.2\times 10^{-3}s}{(150\Omega)(23\times 10^{-6}F)}}]$
This simplifies to:
$q=145.3\times 10^{-6}C$
