In the given diagram, $3\mathit{\Omega }$ and $1.5\mathit{\Omega }$ resistor are in parallel.
$\therefore \frac{1}{{\mathit{R}}_{\mathit{p}}}=\frac{1}{3}+\frac{1}{1\cdot 5}$
$\frac{1}{{\mathit{R}}_{\mathit{p}}}=\frac{1+2}{3}=\frac{3}{3}$
$\therefore {\mathit{R}}_{\mathit{p}}=1\mathit{\Omega }$
Now ${\mathit{R}}_{\mathit{p}}$ and $4\mathit{\Omega }$ resistances are in series
$\therefore {\mathit{R}}_{\mathit{s}}=1+4=5\mathit{\Omega }$
So net external resistance of the circuit $=5\mathit{\Omega }$

$\Rightarrow 0\cdot 3=\frac{1\cdot 8}{\left(5+\mathit{r}\right)}$
$\Rightarrow 0.3\left(5+\mathit{r}\right)=1.8$
$\Rightarrow \left(5+\mathit{r}\right)=6$
$\Rightarrow \mathit{r}=6-5$
$\mathit{r}=1\mathit{\Omega }$