We can use the currents or the tension divider, both work out the same.
The current in the 1st branch (the closest to the source) is
I1 = U / Rtot = 12 / (16+8) = 0.5 A
In the 2nd branch it is I2 = U / Rtot = 12 / (8+4) = 1 A
Using a loop with v in it we can write Kirchoff’s voltage law law:
U1 + U2 + v = 0 (the loop with the 16Ohm resistor and the 4Ohm resistor) which gives 16Ohms*0.5mA - 4Ohms*1mA + v = 0
So we get v = - 4 V (in the way the plus and minus where shown on the figure)