A given length of a wire is doubled on itself and this process is repeated once again . by what factor does the resistance of the wire change ( kindly explain how to determine the new area and length )?

Solution:
Resistance, $R=\rho \frac{L}{A}$
Let initial length and area of cross-section be L and A respectively.
So, its volume be $V=L\times A$.
New length and area of the wire be L' and A'.
So, its volume, $V=L\text{'}\times A\text{'}$
According to question,
 $L\text{'}=2\times 2L=4L ...\left(1\right)$   
Volume of the wire remains constant before and after elongation, so:
$$\begin{gathered} V=L\times A=L\text{'}A\text{'} \\ \Rightarrow A\text{'}=\frac{L\times A}{L\text{'}}=\frac{L\times A}{4L}=\frac{A}{4} ...\left(2\right) \end{gathered}$$
So, ratio of initial and final resistance:

$$\begin{gathered} \frac{R}{R\text{'}}=\frac{L}{L\text{'}}\times \frac{A\text{'}}{A} \left(\mathit{\because }\rho = \mathrm{constant}\right) \\ \Rightarrow \frac{R}{R\text{'}}=\frac{L}{L\text{'}}\times \frac{A\text{'}}{A} \\ =\frac{L}{\left(4L\right)}\times \frac{\left({\displaystyle \frac{A}{4}}\right)}{A} \\ =\frac{1}{16} \\ \Rightarrow R\text{'}=16R \end{gathered}$$

So, new resistance will be 16 times the initial resistance.