Answer
$q_3 = 750~\mu C$
Work Step by Step
We can find the equivalent capacitance of $C_1$ and $C_2$ which are in parallel:
$C_{12} = C_1+C_2 = 15.0~\mu F$
We can find the equivalent capacitance of the three capacitors:
$\frac{1}{C_{eq}} = \frac{1}{C_{12}}+\frac{1}{C_3}$
$\frac{1}{C_{eq}} = \frac{1}{15.0~\mu F}+\frac{1}{15.0~\mu F}$
$C_{eq} = 7.50~\mu F$
We can find the charge $q$:
$q = C_{eq}~V = (7.50~\mu F)(100~V) = 750~\mu C$
Note that capacitor 3 will store all of this charge since it is in series with the other two capacitors.
Therefore:
$q_3 = 750~\mu C$
