Answer
$C_2 = 20~\mu F$
Work Step by Step
We can find the equivalent capacitance $C_{eq}$:
$C_{eq} = \frac{Q}{V}$
$C_{eq} = \frac{450\times 10^{-6}~C}{60~V}$
$C_{eq} = 7.5\times 10^{-6}~F$
We can find the capacitance $C_2$ of capacitor 2:
$\frac{1}{C_{eq}} = \frac{1}{C_1}+\frac{1}{C_2}$
$\frac{1}{C_2} = \frac{1}{C_{eq}}-\frac{1}{C_1}$
$\frac{1}{C_2} = \frac{1}{7.5\times 10^{-6}~F}-\frac{1}{12\times 10^{-6}~F}$
$\frac{1}{C_2} = \frac{8}{60\times 10^{-6}~F}-\frac{5}{60\times 10^{-6}~F}$
$\frac{1}{C_2} = \frac{3}{60\times 10^{-6}~F}$
$C_2 = \frac{60\times 10^{-6}~F}{3}$
$C_2 = 20\times 10^{-6}~F$
$C_2 = 20~\mu F$
