Answer
$I=\rm0.415\; A$
from left to right
Work Step by Step
We have two batteries, so let's assume that the greater battery voltage charges the other battery.
So the current comes out from the 15-V battery toward the 9-V battery.
See the figure below.
All of the directions of the current are assumptions, it may not be the right assumptions and we will know the right ones from the signs of the currents' magnitudes.
But there is some sense here. As an example, the two resistors in the middle are in parallel, so the current that passes through the 24-$\Omega$ resistor must be less than that passes through the 12-$\Omega$ resistor. Hence, $I_2\gt I_3$
Also, the sum of the two currents $I_1+I_2=I_4$ is greater than $I_3$. That's why we assumed that the direction of the current in the wire is to the right.
So now let's test our assumptions by applying Kirchoff's laws (loop and junction);
We have 3 loops here, A, B, and C, as we see in the figure below.
Loop A, starting from the lower corner (c) and moving counterclockwise.
$$\Delta V_A=15-10I-24I_3=0$$
$$10I+24I_3=15\tag 1$$
Loop B, starting from the right lower corner and moving counterclockwise.
$$\Delta V_B= 24I_3-12I_2=0$$
$$I_2= 2I_3\tag 2$$
Loop C, starting from the lower corner and moving clockwise.
$$\Delta V_C= 9+6I_1-12I_2 =0$$
$$ 4I_2-2I_1=3\tag 3$$
From the junction point (b);
$$ I_1+I_2=I_4\tag 4$$
From the junction point (c);
$$ I_3+I_4=I\tag 5$$
Now we have 5 unknowns and 5 equations but we need to focus on $I_4$,
From (1), $I=1.5-2.4I_3$, plug that into (5),
$$ I_3+I_4=1.5-2.4I_3 $$
$$ I_4=1.5-3.4I_3 \tag A$$
From (3),
$$I_1=2I_2-1.5$$
where from (2), $I_2=2I_3$
$$I_1=4I_3-1.5$$
Plug $I_1$ from this previous formula into (4),
$$4I_3-1.5+I_2=I_4$$
Hence,
$$4I_3-1.5+2I_3=I_4$$
$$6I_3-1.5 =I_4\tag B$$
From (A) and (B),
$$6I_3-1.5 =1.5-3.4I_3 $$
$$ I_3 =\bf \dfrac{3}{9.4}\;\rm A$$
Plug into (A),
$$ I_4=1.5-3.4 \left[ \dfrac{3}{9.4}\right]=\color{red}{\bf 0.415}\;\rm A $$
The result is positive, so there is a current of 0.415 A flows from left to right, as we expected, through the bottom wire.
