Explanation

(c) In the steady state, no current flows through the branch containing the capacitor. So, the equivalent circuit will be of the form as shown below:
The effective resistance of the circuit is
$R = \frac{2 \times 3}{2 + 3} + 2.8 = 1.2 + 2.8 = 4\Omega$
The current through the circuit is
$i = \frac{E}{R} = \frac{6}{4} = 1.5A$
Let current $i_{1}$ flows through 2$\Omega$ resistance.
$\therefore\ \ \ 2 \times i_{1} = \left( i - i_{1} \right) \times 3$
$\Longrightarrow \ \ 2i_{1} = \left( 1.5 - i_{1} \right) \times 3$
$\Longrightarrow 2i_{1} = 4.5 - 3i_{1}$
$\Longrightarrow 5i_{1} = 4.5$
$\Longrightarrow i_{1} = 0.9A$