Answer
$R_2 = 4000~\Omega$
Work Step by Step
On the graph, when $R_3 = 0$, the current is $6.0~mA$
In this case, $R_3$ acts like an ordinary wire so none of the current will flow through $R_2$.
We can find $R_1$:
$R_1 = \frac{\mathscr{E}}{i}$
$R_1 = \frac{12~V}{6.0~mA}$
$R_1 = 2000~\Omega$
As $R_3 \to \infty$, none of the current will flow through $R_3$; all of the current will flow through $R_1$ and then $R_2$
We can find $R_2$:
$R_1+R_2 = \frac{\mathscr{E}}{i}$
$R_1+R_2 = \frac{12~V}{2.0~mA}$
$R_1+R_2 = 6000~\Omega$
$R_2 = 6000~\Omega - 2000~\Omega$
$R_2 = 4000~\Omega$
