The given circuit in the question is a case of a balanced Wheatstone bridge. Therefore, the circuit can be simplified as given below.

The equivalent resistance for the given circuit will be,

$R_{\mathrm{eq}}=\frac{\left(3+3\right)\times \left(6+6\right)}{\left(3+3\right)+\left(6+6\right)}+2=6 \mathrm{\Omega }$

Therefore, the current supplied by the battery is

$i=\frac{V}{R_{eq}}=\frac{6}{6}=1 \mathrm{A}$