Let the wire 1 is of length = l and wire 2 of length = ${l \over 3}$

Now area of wire 1 = A while area of wire 2 = 3A

For wire 1,
$\Delta$l = $\left( {{F \over {AY}}} \right)l$ ....(i)

For wire 2, let F' force is applied

$\Delta$l = $\left( {{{F'} \over {3AY}}} \right){l \over 3}$ ....(2)

From equation (i) & (ii)

$\left( {{F \over {AY}}} \right)l$ = $\left( {{{F'} \over {3AY}}} \right){l \over 3}$

$\Rightarrow$ F' = 9F