Answer
$1.0\times10^{-4}C, 4.1\times10^{-4}C$.
Work Step by Step
After a very long time, the capacitors are fully charged, and current runs only through the resistors. Find the voltage across the $3.3k\Omega$ resistor by using the left loop.
$$12.0V-I(4600\Omega)=0$$
$$I=\frac{12.0V}{4600\Omega}=2.609\times 10^{-3}A$$
$$V_{3.3k\Omega}=(2.609\times 10^{-3}A)(3300\Omega)=8.609V$$
This resistor is in parallel with both capacitors, so the voltage across each of them is also 8.609 volts. Find the charge on each capacitor.
$$Q_{12\mu F}=CV=(12\times10^{-6}F)(8.609V)=1.0\times10^{-4}C$$
$$Q_{48\mu F}=CV=(48\times10^{-6}F)(8.609V)=4.1\times10^{-4}C$$
