Let the tension on cable A be TA and that on cable B be TB
According to the vertical equilibrium condition,
TA + TB = Mg
As the Cable B is attached at the far right end of the bar, we can eliminate it from the calculation.
Now there are are two forces that remains in the picture out of which one is a weight that produces counterclockwise rotation and other one is tension on cable A which produces clockwise rotation.
Due to these two forces bar is in equilibroim position and hence summation of these torques must be equalto zero.
TA × (3L/4) = Mg × (L/2)
TA = (4/6) Mg = 2/3 Mg
Correct option is Option (C)