The two rightmost resistors are equivalent to a 2Ω resistor, so in parallel with a 2Ω that gives a 1Ω resistor. We can go on like this by reducing the right part and at the end we have only a small circuit with a 2Ω resistor and a voltage source.
We then have the current in that branch equals to i1 = U/R = 12/2 = 6 A
After, we have to use the current divider principle and for each new loop, as we have a 2Ω resistor, we divide the current by two.
Thus the current I is such that I = (6/2)/2 = 1.5 A