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 $(ii),$ 

$\left( {\frac{F}{{AY}}} \right)3l = \left( {\frac{{F'}}{{3AY}}} \right)l \Rightarrow F' = 9F$