Balanced wheat stone bridge in circuit so there is no current in $5 \Omega$ resistor so it can be removed from the circuit.
$ R _{ eq }=\frac{6 \times 12}{6+12}+2$
$ =\frac{6 \times 12}{18}+2 $
$R _{ eq }=6 \Omega $
$ I =\frac{ V }{ R _{ eq }}=\frac{6}{6}=1 Amp$