Okay so here's the approach I took:
The potential difference in each of the circuits must be the same so if we derive equations for both the potential differences we can set them equal to each other and solve for R1:
In the first circuit
V = 2.2(R1)
In the second we have to find the equivalent resistor, since they are connected in series:
1/R1 + 1/R2 + 1/R3... = Rt
We have R2 so...
1/R1 + 1/3.1 = Rt
1/R1 + 0.323 = Rt
So...
V = 1.4(1/R1 + 0.323)
Set those equal:
2.2R1 = 1.4(1/R1 + 0.323)
2.2R1 = 1.4(1/R1) + 0.4522
Now multiply everything by R1 so we can combine like terms:
2.2R1^2 = 1.4 + 0.4522R1
Isolate to form a quadratic
2.2R1^2 - 0.4522R1 - 1.4 = 0
Solving this quadratic:
R1 = 0.90708 or R1 = -0.701
Since R cannot be negative
R1 = 0.907 ohms