Answer
The charge on capacitor 1 is $~~20~\mu C$
Work Step by Step
When capacitor 1 first reaches equilibrium, the potential difference across the capacitor is $10~V$. We can find the charge on capacitor 1:
$q = C_1~V_1 = (10~\mu F)(10~V) = 100~\mu C$
Note that the equivalent capacitance of the three capacitors in parallel is $C_{eq} = 50~\mu F$
After the switch is thrown to the right, we can find the potential difference across each capacitor:
$V = \frac{q}{C_{eq}} = \frac{100~\mu C}{50~\mu F} = 2 ~V$
We can find the charge on capacitor 1:
$q_1 = C_1~V_1 = (10~\mu F)(2~V) = 20~\mu C$
The charge on capacitor 1 is $~~20~\mu C$.
