Please refer the image attached herewith and kindly solve the subdivision (b). Thank you !

Dear Student,
Let the initial length and area of cross-section of the wire be L and A.
Let the resistivity of the material of the wire be $\rho$.
The initial resistance of the wire will then be given by:
$R=\frac{\rho L}{A}$
Given,
the length of the wire is doubled by pulling it. Then its new length will be L' = 2L.
Since no new material is adding, the volume of the material remains constant before and after pulling it.
Let its new area of cr0ss-section after pulling be A'.
Then
$$\begin{gathered} L\text{'}A\text{'} = LA \\ So, \\ A\text{'}=\frac{LA}{L\text{'}}=\frac{LA}{2L}=\frac{A}{2} \end{gathered}$$

Let the r​​​​esistance of the wire after pulling be R'
Then
$R\text{'}=\frac{\rho A\text{'}}{L\text{'}}=\frac{\rho A}{2\times 2L}=\frac{1}{4}.\frac{\rho A}{L}=\frac{R}{4}$
Regards