Answer
The charge on capacitor 1 is $~~100~\mu C$
Work Step by Step
There is a loop in the circuit that includes only the battery and $C_1$
Therefore the potential difference across capacitor 1 is $10.0~V$
We can find the charge on capacitor 1:
$q_1 = C_1~V_1$
$q_1 = (10.0~\mu F)(10.0~V)$
$q_1 = 100~\mu C$
The charge on capacitor 1 is $~~100~\mu C$
