Under the same load, wire A having length 5 . 0m and cross section...

Elongation in uniform wire under load is given by, $\Deltal = \dfrac{Fl}{SY},$ where $S$ is area of cross-section.

As $F$ is same for both wire and $\Deltal$ is also same, we can write

$\dfrac{\Deltal}{F} = \dfrac{l}{SY} ⇒ \dfrac{{l}_{A}}{{S}_{A}{Y}_{A}} = \dfrac{{l}_{B}}{{S}_{B}{Y}_{B}}$

$⇒ \dfrac{5}{\left(2 . 5 \times {10}^{- 5}\right) \times {Y}_{A}} = \dfrac{6}{\left(3 \times {10}^{- 5}\right) \times {Y}_{B}}$

$⇒ \dfrac{{Y}_{A}}{{Y}_{B}} = 1$