Answer
$C_1 = 2.0~\mu F$
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 1 and capacitor 3 are in series, they have the same charge $q$.
We can find $C_1$:
$q_1 = q_3$
$C_1V_1 = C_3V_3$
$C_1 = \frac{C_3V_3}{V_1}$
$C_1 = \frac{5~V}{2~V}~(0.80~\mu F)$
$C_1 = 2.0~\mu F$
