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 \((ii),\) 

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