(b) In series, current,
$i_{1} = \frac{2E}{2 + 2r}$
In parallel, current,
$i_{2} = \frac{E}{2 + \frac{r}{2}} = \frac{2E}{4 + r}$
According to the question
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i_{1} = i_{2}$
$\Longrightarrow \ \ \frac{2E}{4 + r} = \frac{2E}{2 + 2r}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ r = 2\Omega$