d
Let $x$ be resistance per unit length of the wire. Then,
The resistance of the upper portion is
$R_{1}=x l_{1}$
The resistance of the lower portion is
$R_{2}=x l_{2}$
Equivalent resistance between $A$ and $B$ is
$R = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \frac{{\left( {x{l_1}} \right)(x{l_2})}}{{x{l_1} + x{l_2}}}$
$\frac{8}{3}=\frac{x l_{1} l_{2}}{l_{1}+l_{2}}$ or $\frac{8}{3}=\frac{x l_{1} l_{2}}{l_{2}\left(\frac{l_{1}}{l_{2}}+1\right)}$ or $\frac{8}{3}=\frac{x_{1}}{\left(\frac{l_{1}}{l_{2}}+1\right)}$ ......$(i)$
Also $\quad R_{0}=x l_{1}+x l_{2}$
${12=x\left(l_{1}+l_{2}\right)}$
${12=x l_{2}\left(\frac{l_{1}}{l_{2}}+1\right)}$ ....$(ii)$
Divide $(i)$ by $(ii),$ we get
$\frac{{\frac{8}{3}}}{{12}} = \frac{{\frac{{x{l_1}}}{{\left( {\frac{{{l_1}}}{{{l_2}}} + 1} \right)}}}}{{x{l_2}\left( {\frac{{{l_1}}}{{{l_2}}} + 1} \right)}}{\rm{ or }}\frac{8}{{36}} = \frac{{{l_1}}}{{{l_2}{{\left( {\frac{{{l_1}}}{{{l_2}}} + 1} \right)}^2}}}$
$\left(\frac{l_{1}}{l_{2}}+1\right)^{2} \frac{8}{36}=\frac{l_{1}}{l_{2}}$ or $\left(\frac{l_{1}}{l_{2}}+1\right)^{2} \frac{2}{9}=\frac{l_{1}}{l_{2}}$
Let $y=\frac{l_{1}}{l_{2}}$
$\therefore$ $2(y+1)^{2}=9 y$ or $2 y^{2}+2+4 y=9 y$
or $ 2 y^{2}-5 y+2=0$
Solving this quadratic equation, we get
$y=\frac{1}{2}$ or $2 \therefore \frac{l_{1}}{l_{2}}=\frac{1}{2}$