In the circuit shown, potential difference across the $4\mu F$ capacitor is

Equivalent capacitance 

$C_e=\frac{3\times \left(4+2\right)}{3+\left(4+2\right)}\mu F=2\mu F$

$\therefore$ Charge on $3\mu F$ capacitor, $Q=CV=2\times 10\mu C$

$\therefore$Potential difference across $3\mu F$ capacitor $=\frac{2\times 10}{3}\text{ volt = 6.66 volt}$

$\therefore$ Potential difference across $4\mu F$ capacitor $=\left(10-6.66\right)\text{ volt = 3.34 volt}$