Answer
The current in resistance 2 is $~~1.0~mA$
Work Step by Step
After a long time, the capacitor acts like a broken wire so no current flows around the branch with the capacitor; all the current flows through $R_2$ and then $R_1$
We can find the current in $R_2$ and $R_1$:
$i = \frac{\mathscr{E}}{R_1+R_2}$
$i = \frac{30~V}{20~k \Omega+10~k \Omega}$
$i = 1.0\times 10^{-3}~A$
The current in resistance 2 is $~~1.0~mA$
