$ \mathrm{R}_{\text {eq }}=9 \Omega $

$ \mathrm{i}=\frac{9 \mathrm{~V}}{9 \Omega}=1 \mathrm{~A} $

$ \Delta \mathrm{V}_{6 \Omega}=1 \times 6=6 \mathrm{~V} $

$ \mathrm{~V}_{\mathrm{A}}=3 \mathrm{~V}$

So, potential difference across $6 \mu \mathrm{F}$ is $6 \mathrm{~V}$.

Hence

$Q =C \Delta V $

$ =6 \times 6 \times 10^{-6} \mathrm{C} $

$ =36 \mu \mathrm{C}$