In this problem we can consider the three currents i0, i1 and i2 as drawn bellow.
We have i1 + i2 = 1 mA
And the KVL applied for the two loops:
250*i0 = - 125*(i0 – i1) + 250 *(i2 – i0)
125*(i0 + i1) + 500*i1 = 250* (i2 – i0)
We have 3 equations and 3 unknowns thus we can solve the system.
We get i0 = 0.0870. 10-3 A
i1 = 0.05217. 10-3 A
i2 = 0.9783. 10-3 A
Thus the voltage v we are looking for is v = 250*i0 = 21.7 mV