Answer
$C_2 = 0.80~\mu F$
Work Step by Step
Since the potential difference across both branches of the circuit is equal, the potential difference across both branches is $12~V$
Then the potential difference across capacitor 3 is $~~5~V$
Since capacitor 2 and capacitor 3 are in series, they have the same charge $q$.
We can find $C_2$:
$q_2 = q_3$
$C_2V_2 = C_3V_3$
$C_2 = \frac{C_3V_3}{V_2}$
$C_2 = \frac{5~V}{5~V}~(0.80~\mu F)$
$C_2 = 0.80~\mu F$
