A particular force F applied on a wire increases its length by 2 ×...

$Y = \dfrac{\dfrac{F}{A}}{\dfrac{l}{L}} = \dfrac{F \times L}{A \times l}$ [$l$ = Change in length, $L$ = Original length] (where $Y$ is Young's modulus of elasticity)

Since $Y$, $L$ and $A$ remain same.
$⇒ F ∝ l$
$⇒ \dfrac{{F}_{1}}{{F}_{2}} = \dfrac{{l}_{1}}{{l}_{2}}$
$⇒ \dfrac{F}{{F}_{2}} = \dfrac{2 \times {10}^{- 3}}{4 \times {10}^{- 3}}$

$⇒ {F}_{2} = 2F ⇒ {F}_{2} = nF ⇒ n = 2$