The current flowing through R _ 2 is :

As can be seen from the first circuit above, $CB \text{ and } AB$ are in series which is in parallel with $CA$. So, the net resistance is

$R ' = \dfrac{6}{2}Ω = 3Ω$ as shown in the second diagram.

From the second diagram, $DC \text{ and } CA$ is in series which is parallel to $DA$. Hence the resistance is

R ' ' = \left(\dfrac{6 \times 3}{( 6 + 3 )}\right)Ω = 2Ω as shown in third diagram.

From third diagram, both $ED \text{ and } DA$ are in series which is parallel to $EA$.

The equivalent resistance as can be seen from the above simplified circuit is ${R}_{eq} = 2Ω$.

The current $I$ through battery, $I =$ $\dfrac{8}{2} = 4A$.

From the second diagram, the part $EDACD$ has a resistance of $4Ω$, the total current will divide equally between $EDACD \text{ and } EA$. So consider ${I}_{1} = 2A$ flowing through $ED$. Note that ${I}_{1}$ will be the same current flowing in the first diagram through $ED$.

From first diagram, when the resistances are in parallel the current is divided in the inverse ratio. Thus,
$I '$ through $CD = \dfrac{3}{9} \times 2 = \dfrac{2}{3}A$
I ' ' through ${R}_{2} = \dfrac{1}{2}\left(\dfrac{2}{3}\right) = \dfrac{1}{3}A$