I think I did not make my meaning clear. The actual direction of the current (irrespective of what you initially choose in the diagram) and the numerical value of the current do not depend on how you decide to traverse the circuit.
If we go round the circuit as shown in the diagram, keeping the direction of current also as indicated in the diagram, and use standard loop law:
-5V - I (6Ω) + 3V - I (4Ω) = 0, giving a current of - 0.2 A, as Neon32 got. As Kuruman points out, the negative sign simply means that the actual current is opposite to what was shown in the diagram. so the actual current is going counterclockwise round the circuit
On the other hand, keeping the current as it is in the diagram, we can traverse the circuit in the anticlockwise direction, and get:
+5V + I (6Ω) - 3V + I (4Ω) = 0, giving a current again of - 0.2 A, with again the same interpretation of the negative sign. This is what I mean when I said the answer does not depend on the sense of traversing the circuit.
You're right. I misunderstood what "traversing" meant. Indeed, if you traverse the circuit clockwise or counterclockwise, as long as you assume that the current through the 4 Ω resistor is to the right as indicated by the arrow you drew, then you will get a negative value for the current. The negative sign means that the current through the 4 Ω resistor flows to the left. You can see why this is because the right side of that resistor is connected to the positive terminal of a 5 V battery while the left side is connected to the positive terminal of a 3 V battery. Thus, the right side is at higher potential than the left and current flows right to left regardless oh how you draw the arrow or traverse the circuit.
Absolutely. And nothing prevents you from assuming, to start with, that the current is from right to left in the 4 Ω resistor (anticlockwise round the loop), and then, whichever way you traverse the loop, you will get a positive current, as you should.