For the circuit shown below, the current IAB is

The given circuit:

Applying KCL at node A, we get:

V_A - 50	+	V_A - 100	+	V_A - V_B	= 0...............(i)
5kΩ		20kΩ		4kΩ

KCL at node B, we get:

V_B - 30	+	V_B - 60	+	V_B - V_A	= 0...............(ii)
3kΩ		6kΩ		4kΩ

Simplifying equation (i) and (ii), we get:
2V_A – V_B = 20 ….........(iii)
and
V = 3V_B . . . . . …(iv)
V_B = 4V and V_A = 12V
Now,

I_AB =	V_A - V_B	=	12 - 4	= 2mA
	4kΩ		4kΩ

Hence alternative (A) is the correct choice.