The given circuit can be drawn as shown below
Equivalent resistance of circuit,
$R_{e q}=R_{1}+\frac{R_{2} R_{3}}{R_{3}+R_{2}}=3+\frac{2 \times 2}{2+2}=4 \Omega$
Net current drawn from battery
$I=\frac{V}{R}=\frac{8}{4}=2\, A$
Now the current $I_{2}$
$I_{2}=\frac{I R_{3}}{R_{2}+R_{3}}=\frac{2 \times 2}{2+2}=1\, A$
and potential drop across $R_{2}$
$V_{R_{2}}=I_{2} R_{2}=1 \times 2=2\, V$
