In the arrangement shown, the magnitude of each resistance is 1Ω. The equivalent resistance between O and A is given by

$\frac{3}{4} \Omega$

Both B and D are symmetrically located with respect to points O. Hence, the figure can be folded as shown in figure.

The given circuit forms Wheatstone bridge.

∴  $R_{S}=\frac{1+1}{2}=\frac{3}{2} \Omega$

Resistance, $R_{\text {net }}=\frac{\frac{3}{2} × \frac{3}{2}}{\frac{3}{2}+\frac{3}{2}}=\frac{3}{4} \Omega$