For me the question is a bit ambiguous, but if they are saying that the open-circuit battery voltage is 12.0V and the internal resistance is 0.2 Ohms, then your answer is correct, but you have labeled the current direction and voltage drop across the internal resistor backwards. The positive terminal of the battery is "a" on the left, and positive current flows out that end of the battery (so it returns into "b" from the outside direction, right-to-left).
