Given, $\frac{l_{1}}{l_{2}}=a, \frac{r_{1}}{r_{2}}=b, \frac{Y_{1}}{Y_{2}}=c$
Free body diagram of the two blocks brass and steel are


Let Young's modulus of steel is $Y_{1}$ and of brass is $Y_{2}$.
$\therefore Y_{1}=\frac{F_{1} \cdot l_{1}}{A_{1} \cdot \Delta l_{1}}...(i)$
and $Y_{2}=\frac{F_{2} \cdot l_{2}}{A_{2} \cdot \Delta l_{2}}...(ii)$
Dividing Eq. (i) by(ii),
$$
\begin{array}{l}
\frac{Y_{1}}{Y_{2}}=\frac{\frac{F_{1} \cdot l_{1}}{A_{1} \cdot \Delta l_{1}}}{\frac{F_{2} \cdot l_{2}}{A_{2} \cdot \Delta l_{2}}} \\
\text { or } \frac{Y_{1}}{Y_{2}}=\frac{F_{1} \cdot A_{2} \cdot l_{1} \cdot \Delta l_{2}}{F_{2} \cdot A_{1} \cdot l_{2} \cdot \Delta l_{1}}...(ii)
\end{array}
$$
Force on steel wire from free body diagram $T=F_{1}=(2 g)$ newton
Force on brass wire from free body diagram $F_{2}=T^{\prime}=T+2 g=(4 g)$ newton
Now, putting the value of $F_{1}, F_{2}$, in Eq. (iii), we get
$$
\frac{Y_{1}}{Y_{2}}=\left(\frac{2 g}{4 g}\right) \cdot\left(\frac{\pi r_{2}^{2}}{\pi r_{1}^{2}}\right) \cdot\left[\frac{l_{1}}{l_{2}}\right] \cdot\left(\frac{\Delta l_{2}}{\Delta l_{1}}\right)
$$
or $c=\frac{1}{2}\left(\frac{1}{b^{2}}\right) \cdot a\left(\frac{\Delta l_{2}}{\Delta l_{1}}\right)$
or $\frac{\Delta l_{1}}{\Delta l_{2}}=\left(\frac{a}{2 b^{2} c}\right)$