Under steady state condition the potential difference across the capacitor in the circuit is________  V.

Under steady-state DC conditions, the capacitor behaves as an open circuit.

Therefore, no current flows through the middle branch containing the capacitor and the $$4\,\Omega$$ resistor.

Hence, the current flows only through the outer loop containing $$2\,\Omega$$ and $$6\,\Omega$$ resistors and the $$2\,V$$ battery.

Let the current in the outer loop be $$I$$.

Applying Kirchhoff's voltage law,

$$2I+6I=2$$

$$8I=2$$

$$I=0.25\,A$$

The potential difference between the left and right nodes is the potential drop across the $$2\,\Omega$$ resistor:  $$V=IR$$

$$V=0.25\times2$$

$$V=0.5\,V$$

Since no current flows through the $$4\,\Omega$$ resistor, there is no potential drop across it. Therefore, the potential difference across the capacitor is the same as the potential difference between the two nodes.

Thus,  $$V_C=0.5\,V$$

Hence, the correct option is A.