You will apply Kirchoff's rules to solve this problem.
Define the direction of flow in each of the two loops. We'll define the left loop as clockwise and the right loop as counterclockwise. I chose these directions because intuitively, those appear to be correct, but the fact is that you could define both as clockwise, too. It doesn't matter; if you choose the wrong direction, you'll just get a negative value for current.
Set up your equations for each loop:
E1 – i1r1 – i3r3 = 0. [Kirchoff's loop rule, eq.1]
E2 – i2r2 – i3r3 = 0. [Kirchoff's loop rule, eq.2]
i1 + i2 = i3. [Kirchoff's junction rule, eq.3]
Plug in values and solve the system for your unknowns:
9 – 3i1 – 6 = 0 [eq.1]
3i1 = 3
i1 = 1
1 + i2 = 3 [eq.3]
i2 = 2
E2 – 9 – 6 = 0 [eq.2]
E2 = 15
So you have 1.0A through the 3Ω resistor, 2.0A through the 4.5Ω resistor, and the second battery is 15V.