We model the cell's internal resistance as an additional resistor in the circuit. We will use V = IR to calculate the voltage drops over each resistor for both circuits.
Circuit 1: 0.9A current, internal resistance r, resistor 2Ω
V = 0.9r + 0.9(2) = 0.9r + 1.8
Circuit 2: 0.3A current, internal resistance r, resistor 7Ω
V = 0.3r + 0.3(7) = 0.3r + 2.1
Since V should be the same for both circuits, set the total voltage drops equal to each other and solve for r:
0.9r +1.8 = 0.3r + 2.1
0.6r = 0.3
r = 0.5Ω
