Since initial volume of wires are same
$\therefore $ Their areas of cross sections are $A$ and $3A$ and lengths are $3l$ and $l$ respectively.
For Wire $1,$
$\Delta l = \left( {\frac{F}{{AY}}} \right)3l\,$ $...(i)$
For wire $2$, let $F'$ force is applied
$\frac{{F'}}{{3A}} = Y\frac{{\Delta l}}{l}$
$ \Rightarrow \Delta l = \left( {\frac{{F'}}{{3AY}}} \right)l$
From eqns $(i)$ and $(ii),$
$\left( {\frac{F}{{AY}}} \right)3l = \left( {\frac{{F'}}{{3AY}}} \right)l \Rightarrow F' = 9F$
