We can simplify the network as shown :
So, net resistance,
$R=24+1.6=4.0 \Omega$
Therefore, current from the battery,
$i=\frac{V}{R}=\frac{4}{4}=1 A$
Now from the circuit (b),
$4 I'=6 I$
$\Rightarrow I'=\frac{3}{2}I$
but $i=I+I'=I+\frac{3}{2} I=\frac{5}{2} I$
$\therefore 1=\frac{5}{2} I $
$\Rightarrow I=\frac{2}{5}=0.4 \,A$
