Answer
0.90 coulombs
Work Step by Step
The work needed to move the q=0.30mC charge between the two capacitor plates is W=qV, where V is the voltage difference between the plates due to the charge Q already on them, $V=\frac{Q}{C}$.
Assume that q is much less than Q, so that the charge on the capacitor doesn’t change by much.
$$W=qV=q\frac{Q}{C}$$
$$Q=\frac{CW}{q}=\frac{(15\times10^{-6}F)(18J)}{0.30\times10^{-3}C}=0.90C$$
It is true that q is much less than Q.
