Answer
$i_1 = 1.1~mA$
Work Step by Step
Initially, the capacitor acts like an ordinary wire.
We can find the equivalent resistance of $R_2$ and $R_3$ which are in parallel:
$\frac{1}{R_{23}} = \frac{1}{0.73~M \Omega}+\frac{1}{0.73~M \Omega}$
$R_{23} = 0.365~M \Omega$
We can find the equivalent resistance of the circuit:
$R_{eq} = 0.73~M \Omega+0.365~M \Omega = 1.095~M \Omega$
We can find the current in the circuit:
$i = \frac{1.2~kV}{1.095~M \Omega} = 1.1~mA$
Since all of the current flows through $R_1$, then $~~i_1 = 1.1~mA$
