Answer
$4R$
Work Step by Step
We know that the resistance is given by
$$R=\dfrac{\rho L}{A}=\dfrac{\rho L}{\pi r^2}$$
Hence, the initial resistance of the wire is
$$R =\dfrac{\rho L_i}{\pi r_i^2}\tag 1$$
And when it is stretched, the final resistance is
$$R_f=\dfrac{\rho L_f}{\pi r_f^2}$$
where the wire is stretched to twice its initial length, so $L_f=2L_i$
$$R_f=\dfrac{2\rho L_i}{\pi r_f^2}\tag 2$$
We know that the volume of the wire is constant so the wire has the same volume in both cases, so
$$V_i=V_f$$
$$A_iL_i=A_fL_f$$
$$ \color{red}{\bf\not} \pi r_i^2 \color{red}{\bf\not} L_i=2 \color{red}{\bf\not} \pi r_f^2 \color{red}{\bf\not} L_i $$
Hence,
$$r_f^2=\frac{1}{2}r_i^2$$
Plug into (2);
$$R_f=\dfrac{4\rho L_i}{\pi r_i^2} =4\dfrac{ \rho L_i}{\pi r_i^2}$$
where $=\dfrac{ \rho L_i}{\pi r_i^2}=R$, so
$$\boxed{R_f =4R}$$
