Answer
$C_3 = 2.0~\mu F$
Work Step by Step
Since $8.0~\mu C$ passes through point b, then the charge on capacitor 3 must be $4.0~\mu C$
Note that the potential difference across capacitor 3 and capacitor 4 must be equal since they are in parallel.
We can find $C_3$:
$V_3 = V_4$
$\frac{q_3}{C_3} = \frac{q_4}{C_4}$
$C_3~q_4 = C_4~q_3$
$C_3 = \frac{q_3}{q_4}~C_4$
$C_3 = (\frac{4.0~\mu C}{8.0~\mu C})~(4.0~\mu F)$
$C_3 = 2.0~\mu F$
