Answer
$q_2 = 480~\mu C$
Work Step by Step
We can find the equivalent capacitance of the capacitors in series:
$\frac{1}{C_{eq}} = \frac{1}{C_1}+\frac{1}{C_2}$
$\frac{1}{C_{eq}} = \frac{1}{2.00~\mu F}+\frac{1}{8.00~\mu F}$
$\frac{1}{C_{eq}} = \frac{4}{8.00~\mu F}+\frac{1}{8.00~\mu F}$
$C_{eq} = \frac{8.00~\mu F}{5}$
$C_{eq} = 1.60~\mu F$
Since the capacitors are in series, the charge on each capacitor is equal.
We can find the charge on each capacitor:
$q = C_{eq}~V = (1.60~\mu F)(300~V) = 480~\mu C$
Note that this is the charge on each capacitor.
Therefore:
$q_2 = 480~\mu C$
