Answer
The charge on capacitor 4 is $~~16~\mu C$
Work Step by Step
The equivalent capacitance of $C_3$ and $C_4$ is $1.0~\mu F+2.0~\mu F = 3.0~\mu F$
The equivalent capacitance of $C_1$ and $C_2$ is $2.0~\mu F+4.0~\mu F = 6.0~\mu F$
We can find the equivalent capacitance of the four capacitors:
$\frac{1}{C_{eq}} = \frac{1}{3.0~\mu F}+\frac{1}{6.0~\mu F}$
$C_{eq} = 2.0~\mu F$
We can find the charge stored on $C_{eq}$:
$q = C_{eq}~V$
$q = (2.0~\mu F)(12~V)$
$q = 24~\mu C$
Then the total charge stored on $C_3$ and $C_4$ is $24~\mu C$
Note that the potential difference across $C_3$ and $C_4$ must be equal.
We can find the charge $q_4$ on capacitor 4:
$V_3 = V_4$
$\frac{q_3}{C_3} = \frac{q_4}{C_4}$
$\frac{24~\mu C-q_4}{C_3} = \frac{q_4}{2~C_3}$
$48~\mu C - 2~q_4 = q_4$
$48~\mu C = 3~q_4$
$q_4 = 16~\mu C$
The charge on capacitor 4 is $~~16~\mu C$
