Think of the internal resistance as just another resistor in the series. I believe that is referred to as R1 in the problem. Since the total resistance of a series circuit is the sum of the individual resistances, that would be 17.0 Ω. By Ohm's law, the current in the circuit is 1.5 v / 17.0 Ω ≈ 0.088 amperes.
Ohm's law (in form V=IR) is used again to calculate the voltage drop across each individual resistor. For R1,
V = (0.088 A) (3.00 Ω) = 0.26 v
The terminal voltage is 1.5 v - 0.26 v ≈ 1.2 v