The current flowing through R₂ is :

Equivalent resistance $=2\mathrm{\Omega }$
$\Rightarrow$current from battery $=\frac{8}{2}=4\mathrm{A}$
So, 2A flows through R₅
Which splits in 2:1 ratio between R₇ and R₄
Further it splits in 1:1 ratio between R₃ and R₂
So, current through ${\mathrm{R}}_2=2\times \frac{1}{3}\times \frac{1}{2}=\frac{1}{3}$