Answer
$C_1 = 4.0~\mu F$
Work Step by Step
We can find the potential difference across capacitor 2:
$V_2 = \frac{q_2}{C_2} = \frac{12~\mu C}{3.0~\mu F} = 4.0~V$
We can find the potential difference across capacitor 4:
$V_4 = \frac{q_4}{C_4} = \frac{8.0~\mu C}{4.0~\mu F} = 2.0~V$
Since the total potential difference across capacitor 1, capacitor 2, and capacitor 4 is $9.0~V$, the potential difference across capacitor 1 must be $3.0~V$
We can find $C_1$:
$C_1 = \frac{q_1}{V_1} = \frac{12~\mu C}{3.0~V} = 4.0~\mu F$
