The charge on 4 $\mathrm{\mu }$F capacitor in the given circuit is (in $\mathrm{\mu }$C)

The circuit can be redrawn as,

Here, 4 $\mathrm{\mu }$F and 6 $\mathrm{\mu }$F are in series. So, charge is same on both.
Now, equivalent capacity between A and B,

                                    ${\mathrm{C}}_{\mathrm{AB}}=\frac{6\times 4}{6+4}=2.4\mathrm{\mu F}$

So, charge on 4  $\mathrm{\mu }$F capacitor,

                                    $\mathrm{Q}={\mathrm{C}}_{\mathrm{AB}}\times 10=2.4\times 10=24\mathrm{\mu C}$