Answer
At $t = 4.00~ms$, the current in resistor 2 is $~~0.411~mA$
Work Step by Step
When the steady state is reached, note that no current flows in the branch with the capacitor. We can find the current through $R_1$ and $R_2$:
$i = \frac{\mathscr{E}}{R_1+R_2}$
$i = \frac{20.0~V}{10.0~k \Omega+15.0~k \Omega}$
$i = 8.00\times 10^{-4}~A$
We can find the potential difference across $R_2$:
$\Delta V = i~R_2$
$\Delta V = (8.00\times 10^{-4}~A)(15.0~k \Omega)$
$\Delta V = 12.0~V$
Note that at $t=0$, the potential difference across the capacitor will also be $12.0~V$
When the steady state is reached, then the charge on the capacitor is $~~q_0 = (C)~(12.0~V)$
We can find the current in resistor 2 at $t = 4.00~ms$:
$i = (\frac{q_0}{R_2C})~e^{-t/R_2C}$
$i = (\frac{12.0~V}{R_2})~e^{-t/R_2C}$
$i = (\frac{12.0~V}{15.0~k \Omega})~e^{-4.00~ms/(15.0~k \Omega)(0.400~\mu F)}$
$i = (\frac{12.0~V}{15.0~k \Omega})~(0.5134)$
$i = 0.411~mA$
At $t = 4.00~ms$, the current in resistor 2 is $~~0.411~mA$
