Answer
$i_1 = 0.82~mA$
Work Step by Step
After a long time, the capacitor acts like a broken wire, so no current flows through the branch with the capacitor. All the current flows through $R_1$ and $R_2$
We can find the equivalent resistance of $R_1$ and $R_2$:
$R_{eq} = 0.73~M \Omega+0.73~M \Omega = 1.46~M \Omega$
We can find the current in the circuit:
$i = \frac{1.2~kV}{1.46~M \Omega} = 0.82~mA$
$i_1 = 0.82~mA$
