Answer
$V_2 = 400~V$
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 $R_2 = R_3$, half of the current flows through $R_2$ and half of the current flows through $R_3$
We can find $i_2$:
$i_2 = \frac{i}{2} = \frac{1.1~mA}{2} = 0.55~mA$
We can find $V_2$:
$V_2 = i_2~R_2$
$V_2 = (0.55~mA)(0.73~M \Omega)$
$V_2 = 400~V$
