Using the condition of a balanced wheat stone bridge,
$\Rightarrow \frac{ R }{3}=\frac{4}{6} \Rightarrow R =2\, \Omega$
So the effective resistance of the circuit is
$R _{ eq }=\frac{6 \times 9}{6+9}=\frac{18}{5} \Omega$
$i =\frac{36}{ R _{ eq }}=10 \,A$