A 12\ V battery is connected across two parallel branches. The...

At steady state, the capacitor is fully charged and acts as an open circuit.

Therefore, no current flows through the second branch containing the $6\ \Omega$ resistor and the capacitor.

The entire current from the battery flows only through the first branch containing the $4\ \Omega$ resistor.

Using Ohm's law, the steady-state current is:
$I = \dfrac{V}{R} = \dfrac{12}{4} = 3\ \text{A}$

Answer: $3\ \text{A}$