Answer
The final charge on capacitor 2 is $~~16.0~\mu C$
Work Step by Step
When the switch is to the left, we can find the charge on capacitor 1 when it is in equilibrium:
$q = C_1~V = (4.00~\mu F)(12.0~V) = 48.0~\mu C$
We can find the equivalent capacitance of $C_2$ and $C_3$ which are in series:
$\frac{1}{C_{23}} = \frac{1}{6.00~\mu F}+\frac{1}{3.00~\mu F}$
$\frac{1}{C_{23}} = \frac{1}{6.00~\mu F}+\frac{2}{6.00~\mu F}$
$C_{23} = 2.00~\mu F$
After the switch is moved to the right, $C_1$ is in parallel with the other two capacitors.
We can find the equivalent capacitance of the three capacitors:
$C_{eq} = 4.00~\mu F+2.00~\mu F = 6.00~\mu F$
We can find the potential difference between the top of the circuit and the bottom of the circuit:
$V = \frac{q}{C_{eq}} = \frac{48.0~\mu C}{6.00~\mu F} = 8.00~V$
We can find the final charge on capacitor 1:
$q_1 = C_1~V = (4.00~\mu F)(8.00~V) = 32.0~\mu C$
We can find the final charge on capacitor 2:
$q_2 = 48.0~\mu C-32.0~\mu C = 16.0~\mu C$
The final charge on capacitor 2 is $~~16.0~\mu C$
