A battery of $$6 \text{ V}$$ is connected to the circuit as shown below. The current $$I$$ drawn from the battery is

The central part of the circuit is a Wheatstone Bridge ($$\frac{3}{3} = \frac{6}{6}$$)

Removing the central $$5\text{ }\Omega$$ resistor leaves us with two parallel branches:

Top branch: Two $$3\text{ }\Omega$$ resistors in series $$\implies R_{\text{top}} = 3 + 3 = 6\text{ }\Omega$$

Bottom branch: Two $$6\text{ }\Omega$$ resistors in series $$\implies R_{\text{bottom}} = 6 + 6 = 12\text{ }\Omega$$

$$R_{\text{bridge}} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4\text{ }\Omega$$

$$R_{\text{total}} = R_{\text{bridge}} + 2\text{ }\Omega = 4 + 2 = 6\text{ }\Omega$$

$$I = \frac{6\text{ V}}{6\text{ }\Omega} = \mathbf{1\text{ A}}$$