Answer
The current in resistance 1 is $~~1.5~mA$
Work Step by Step
At $t=0$, the capacitor acts like an ordinary wire so all the current flows around the branch with the capacitor; no current flows through $R_2$
We can find the current in $R_1$:
$i = \frac{\mathscr{E}}{R_1}$
$i = \frac{30~V}{20~k \Omega}$
$i = 1.5\times 10^{-3}~A$
The current in resistance 1 is $~~1.5~mA$
