Answer
$q_2 = 250~\mu C$
Work Step by Step
In part (a), we found that $q_3 = 750~\mu C$
Since $C_1$ and $C_2$ are in parallel, the total charge on capacitor 1 and capacitor 2 is $750~\mu C$. Note that $q_1 = q_3-q_2$
Since $C_1$ and $C_2$ are in parallel, the potential difference across capacitor 1 and capacitor 2 is equal.
We can find $q_2$:
$V_2 = V_1$
$\frac{q_2}{C_2} = \frac{q_1}{C_1}$
$q_2~C_1 = C_2~q_1$
$q_2~C_1 = C_2~(q_3-q_2)$
$q_2~(C_1+C_2) = C_2~q_3$
$q_2 = \frac{C_2}{C_1+C_2}~q_3$
$q_2 = (\frac{5.00~\mu F}{15.0~\mu F})~(750~\mu C)$
$q_2 = 250~\mu C$
