Answer
$V_b-V_a = 39V$ or $V_b-V_a =-15V$.
Work Step by Step
There are two answers because the 3.10 mA current could be flowing to the right, or to the left, through the $4.0 k\Omega$ resistor.
First, assume the current is to the right. The voltage across the $4.0 k\Omega$ resistor is given by Ohm’s law as V=IR = 12.4 volts, which is the same as the voltage drop across the $8.0 k\Omega$ resistor. The current through it is half that through the $4.0 k\Omega$ resistor because it has twice the resistance. The total current flowing in the circuit is the sum of the two currents, 4.65 mA.
Using the current 4.65 mA, find the terminal voltage of the battery. Write a loop equation, starting at the negative terminal of the unknown battery and proceeding clockwise.
$$V_{ab}-3200\Omega(4.65mA)-12.4V-12.0V-1.0\Omega(4.65mA)=0$$
$$V_{ab}=24.4V+3201\Omega(4.65mA) \approx39V$$
Now assume the current is to the left. The voltage drop across the parallel combination is still 12.4 V, but now the right hand side is at higher voltage.
Write a loop equation, starting at the negative terminal of the unknown battery and proceeding clockwise. Since the current is now assumed to flow to counterclockwise, we would not be surprised to find that $V_{ab}=V_b-V_a$ is negative, meaning the battery actually points the other way, compared to what is shown.
$$V_{ab}+3200\Omega(4.65mA)+12.4V-12.0V+1.0\Omega(4.65mA)=0$$
$$V_{ab}=-0.4V-3201\Omega(4.65mA) \approx-15V$$
