A wire of length L and radius r is clamped rigidly at one end. When...

Young's modulus is given by $Y = \dfrac{FL}{A ∆ L}$, here, $A$ is cross-section area and $∆ L$ is change in length.

Now, when force $F$ is applied, change in length $\Delta{L}_{1} = \dfrac{FL}{AY} = \dfrac{FL}{\pi{r}^{2}Y} = 5cm$

Now, when force $4F$ is applied, change in length

$\Delta{L}_{2} = \dfrac{4F4L}{\pi16{r}^{2}Y} = \dfrac{FL}{\pi{r}^{2}Y} = 5cm$