Answer
The resistance of resistor 2 is $~~200~\Omega$
Work Step by Step
Let $\Delta V_D$ be the potential difference across resistor 3.
$V_A = 12.0~V$
Therefore:
$\Delta V_B+\Delta V_C+\Delta V_D = 12.0~V$ since the two branches are in parallel.
It is given that $\Delta V_B = 2.00~V$ and $\Delta V_C = 5.00~V$
Then $\Delta V_D = 5.00~V$
We can find the current through the resistor 3:
$i_3 = \frac{\Delta V_D}{R_3} = \frac{5.00~V}{200~\Omega} = \frac{1}{40}~A$
This is also the current through resistor 2 since resistors 1, 2, and 3 are connected in series.
We can find the resistance of resistor 2:
$R_2 = \frac{\Delta V_C}{i} = \frac{5.00~V}{\frac{1}{40}~A} = 200~\Omega$
The resistance of resistor 2 is $~~200~\Omega$.
