Let the current flowing through the circuit be $I \mathrm{A}$ in anti-clockwise direction.

The net resistance is $R=4.5 \mathrm{\Omega }+2 \mathrm{\Omega }+1.5 \mathrm{\Omega }=8 \mathrm{\Omega }$

Hence, $I=\frac{V}{R}=\frac{4 \mathrm{V}}{8 \mathrm{\Omega }}=0.5 \mathrm{A}$

The current flowing through the $3 \mathrm{\Omega }$ resistor is 

$\frac{6}{6+3}\times 0.5 =\frac{1}{3} \mathrm{A}$

So, $x=1$