Let the current through $3\mathit{\Omega }$ resistor be ' $\mathit{i}$ '
A. So the current through $1.5\mathit{\Omega }$ resistor will be $\left(0.3-\mathit{i}\right)$ A. As $3\mathit{\Omega }$ resistor and $1.5\mathit{\Omega }$ resistor are in parallel. So,
$3\times \mathit{i}=1.5\left(0.3-\mathit{i}\right)$
$\Rightarrow 3\mathit{i}=0.45-1.5\mathit{i}$
$\Rightarrow 4.5\mathit{i}=0.45$
$=\frac{0.45}{4.5}=0.1\mathrm{A}$