The upper part of the circuit is balanced Wheat stone's bridge. So, resistance $R$ in branch $C D$ is ineffective as no current will flow in this branch. The circuit therefore reduces to three parallel branches having resistance $R, R+R$ and $R+R$.
$\therefore $ Effective resistance $\frac{1}{R'}=\frac{1}{R}+\frac{1}{2 R}+\frac{1}{2 R}$
$\Rightarrow \, R'=R / 2$