The current flowing through $R_2$ is :

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

$R\text{'}=\frac{6}{2} \mathrm{\Omega }=3 \mathrm{\Omega }$ as shown in the second diagram.

From the second diagram, $DC \& CA$ is in series which is parallel to $DA$. Hence the resistance is

$R\text{'}\text{'}=\left(\frac{6\times 3}{(6+3)}\right) \mathrm{\Omega }=2 \mathrm{\Omega }$ as shown in third diagram.

From third diagram, both $ED \& DA$ are in series which is parallel to $EA$.

The equivalent resistance as can be seen from the above simplified circuit is $R_{eq}=2 \mathrm{\Omega }$.

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

From the second diagram, the part $EDACD$ has a resistance of $4 \mathrm{\Omega }$, the total current will divide equally between $EDACD \& EA$. So consider $I_1=2 \mathrm{A}$ 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,
$I\text{'}$ through $CD=\frac{3}{9}\times 2=\frac{2}{3} \mathrm{A}$
$I\text{'}\text{'}$ through $R_2=\frac{1}{2}\left(\frac{2}{3}\right)=\frac{1}{3} \mathrm{A}$