If current through R-ohm resistor is I amp, AO branch carries the same current, since, current through the branch CO is zero. This also means that the nodes C and O are at the equal potential. Then, equating voltage-drops, we have VAO = VAC.
This means branch AC carries a current of 4I.
This is current of 4 I also flows through the branch CB. Equating the voltage-drops in branches OB and CB,
1.5 × 4 I =R I, giving R = 6 Ω
At node A, applying KCL, a current of 5 I flows through the branch BA from B to A. Applying KVL around the loop BAOB, I = 0.5 Amp.