Two wires are made of the same material and have the same volume....

$Y = \dfrac{Fl}{A ∆ l}$

$∵ V = Alsol = \dfrac{V}{A}$

$F = \dfrac{YA ∆ l}{l} = \dfrac{Y{A}^{2} ∆ l}{V}$

$\dfrac{{F}_{1}}{{F}_{2}} = {\left(\dfrac{{A}_{1}}{{A}_{2}}\right)}^{2} ⇒ \dfrac{F}{{F}_{2}} = {\left(\dfrac{A}{3A}\right)}^{2} = \dfrac{1}{9}$

${F}_{2} = 9{F}_{1} = 9F$