Six equal resistances each of 4 Ω are connected to form a network as…

The correct answer is Option (1) → 2 Ω

AOBC here form a Wheatstone Bridge and therefore, there will be no current in arm CO.

Now,

$\frac{1}{R_{eff}}=\frac{1}{4}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$

$R_{eff}=2$