Again, here is the circuit we have:
We can apply the KVL for the 3 loops by expressing everything in terms of i0, i1 and i2.
We have:
12 = 2* (i1 – i0) + 4*i1
12 = 3*i2 + 2*(i0 + i2)
i0 = 1 A
We have 3 equations for 3 unknowns so we can solve the system. We get
i0 = 1 A
i1 = 2.33 A
i2 = 2 A
Thus the voltage we are looking for is v = - 2* (i1 – i0) + 2*(i0 + i2)
So v = 3.33 V