$3 i _1=1$

$i _1=\frac{1}{3} A$

$V _{ A }-1 \cdot i _1-1= V _{ B }$

$V _{ A }- V _{ B }=1+ i _1=\frac{4}{3} \text { volt }$

Potential drop across capacitor $\Delta V =\frac{4}{3}$ volt

$\therefore \quad$ Charge on capacitor $q _1= C \Delta V$

$= 1 \times \frac{4}{3} \mu C$

$q _1=  1.33 \mu C$

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$V _{ A }-1 \cdot i _2+2-2 i _2= V _{ A }$

$3 i _2=2$

$i _2=\frac{2}{3} A$

$V _{ A }- i _2 \times 1= V _{ B }$

$V _{ A }- V _{ B }= i _2 \times 1=\frac{2}{3} \text { volt }$

Potential difference across capacitor $\Delta V =\frac{2}{3}$ volt

$\therefore \quad \text { Charge on capacitor } q _2  = C \Delta V$

$=1 \times \frac{2}{3}=0.67 \mu C$