I[itex]_{1}[/itex] + 4 mA = 2 mA
I[itex]_{1}[/itex] = -2 mA
This problem looks to have 2 unknowns. You aren't given the current flowing through the resistor on the left between the 2mA and 12mA legs. I would think you would need to write 2 equations to solve for I1 and that unknown current, but I could be wrong.