If r3 is open, the total resistance is 270+560 = 830 ohyms and the current would be 24.6959/830 or about 30 mA. Hence R2 must be open since the measured current is more than 16,8 mA.
R1 is in parallel with R3. Add the Ohms on 1 & 3 = 1470 ohms. Volts devided by ohms = amps. 1470 devided by 24.6954 = .016799 amps or 16.799 ma.
Just do the math and use logic.
If R1 is open, current is zero.
If R2 is open, you have R1 and R3 is series for 1470 ohms and I = 24.7/1470 = 16.8 mA
If R3 is open, you have R1 and R2 in series for 830 ohms and I = 24.7/830 = 29.8 mA
So obviously R2 is open.
PS, voltages and currents to 6 places are silly. 3 is enough.
R = E/I so with the given voltage and current, we know the resistance of the series-parallel combination is roughly 1470 ohms.
The only way you get 1470 ohms from the series parallel combination you have provided is by R2 being open. In that case, the effective resistance becomes R1 + R3 = 1470 ohms (meaning R2 is open)