The give circuit is a balanced Wheatstone's bridge.
In the given circuit, ratio of resistances in the opposite arms is same, hence bridge is a balance one, hence diagonal resistance of $10 \,\Omega$ will be in effective. The circuit now reduces to
$R_{1}=10 \Omega+10 \Omega=20 \Omega $
$R_{2}=10 \Omega+10 \Omega=20\, \Omega$
$\therefore $ Effective resistance is
$\frac{1}{R'}=\frac{1}{20}+\frac{1}{20} $
$\frac{1}{R'} =\frac{40}{20 \times 20}=\frac{1}{10}$
$R''=10 \,\Omega$
