Explanation

(d) Let $x$ is the resistance per unit length then
Equivalent resistance
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$R = \frac{R_{1}R_{2}}{R_{1} + R_{2}}$
$\Rightarrow \frac{8}{3} = \frac{\left( x\mathcal{l}_{1} \right)\left( x\mathcal{l}_{2} \right)}{x\mathcal{l}_{1} + x\mathcal{l}_{2}} \Rightarrow \frac{8}{3} = x\frac{\mathcal{l}_{1}\mathcal{l}_{2}}{\mathcal{l}_{1} + \mathcal{l}_{2}}$
$\Rightarrow \frac{8}{3} = x\frac{\mathcal{l}_{1}}{\frac{\mathcal{l}_{1}}{\mathcal{l}_{2}} + 1}$ …(i)
Also $R_{0} = x\mathcal{l}_{1} + x\mathcal{l}_{2} \Rightarrow 12 = x\left( \mathcal{l}_{1} + \mathcal{l}_{2} \right)$
$\Rightarrow 12 = x\mathcal{l}_{2}\left( \frac{\mathcal{l}_{1}}{\mathcal{l}_{2}} + 1 \right)$ …(ii)
Dividing eq. (i) by eq. (ii)
$\frac{\frac{8}{3}}{\frac{12}{1}} = \frac{\frac{x\mathcal{l}_{1}}{\left( \frac{\mathcal{l}_{1}}{\mathcal{l}_{2}} + 1 \right)}}{x\mathcal{l}_{2}\left( \frac{\mathcal{l}_{1}}{\mathcal{l}_{2}} + 1 \right)} = \frac{\mathcal{l}_{1}}{\mathcal{l}_{2}\left( \frac{\mathcal{l}_{1}}{\mathcal{l}_{2}} + 1 \right)^{2}} \Rightarrow \left( \frac{\mathcal{l}_{1}}{\mathcal{l}_{2}} + 1 \right)^{2} \times \frac{8}{36} = \frac{\mathcal{l}_{1}}{\mathcal{l}_{2}}$
$\Rightarrow \left( y^{2} + 1 + 2y \right) \times \frac{8}{36} = y\ \ \ \ \ \ \ \ \lbrack Where\ y = \frac{\mathcal{l}_{1}}{\mathcal{l}_{2}}\rbrack$
$\Rightarrow 8y^{2} + 8 + 16y = 36y$
$\Rightarrow 8y^{2} - 20y + 8 = 0$
On solving, we get $y = \frac{1}{2}$ or $2$
$\therefore y = \frac{\mathcal{l}_{1}}{\mathcal{l}_{2}} = \frac{1}{2}$ or $2$