c
The current in capacitor is given by,
$i=\frac{9}{27 \times 10^{3}}$
Now, the voltage is given by,
$V_{c}=15 \times 10^{3} \times \frac{9}{27 \times 10^{3}}$
The Charge is given by,
$q=9 \times 10^{-6} \times \frac{15}{3}$
$=45 \times 10^{-6} C$
$=45 \mu C$