The Young's modulus of material of wire 1 ~is twice that of wire 2...

Let us assume,&
Young modulus& \left(1\right)$~be,~${Y}_{1}$<br />Young modulus~ \left(2\right)$~be,~${Y}_{2}$
Length of wire \left(1\right)$~be,~${L}_{1}$<br />Length of wire~ \left(2\right)$~be,~${L}_{2}$
Force on wire be,~$F$
Extension of both wire be~${δ}_{1} \text{ and } {δ}_{2}$

Given
Young's modulus is,~${Y}_{1} = 2{Y}_{2}$
Length of both wire is same,~${L}_{1} = {L}_{2} = L$
Radius of~ \left(1\right)$~be,~${R}_{1}$<br />Radius of~ \left(2\right)$~be,~${R}_{2}$
Forces on both are same.
Extension of both wire be,~${δ}_{1} = {δ}_{2} = \delta$

From Hooke's law, we have~$F = \dfrac{AY\delta}{L}$
For~ \left(1\right)$,~$F = \dfrac{\pi{R1}^{2}{Y}_{1}\delta}{L1} = \dfrac{\pi{R1}^{2} \left(2{Y}_{2}\right)\delta}{L} . . . \left(1\right)<br />For~ \left(2\right)$,~$F = \dfrac{\pi{R2}^{2}{Y}_{2}\delta}{L2} = \dfrac{\pi{R2}^{2}{Y}_{2}\delta}{L} . . . \left(2\right)

Equating~ \left(1\right) \text{ and } \left(2\right)$, we have<br />$⇒ $\dfrac{\pi{R1}^{2}\left(2{Y}_{2}\right)\delta}{L} = \dfrac{\pi{R2}^{2}{Y}_{2}\delta}{L}<br />$⇒ {R1}^{2}\left(2{Y}_{2}\right) = {R2}^{2}{Y}_{2}$<br />$⇒ {2{R}_{1}}^{2} = {R2}^{2}$<br />$⇒ {R}_{1} = \dfrac{R2}{2}