Resistance of a wire,
$R = \rho {l \over A} = 4\Omega$   ...(i)

When wire is stretched twice, its new length be $l'$. Then

$l' = 2l$

On stretching volume of the wire remains constant.

$\therefore lA = l'A'$ where A' is the new cross-sectional area

$\Rightarrow A' = {l \over {l'}}A = {l \over {2l}}A = {A \over 2}$

$\therefore$ Resistance of the stretched wire is

$R' = \rho {{l'} \over {A'}} = \rho {{2l} \over {\left( {A/2} \right)}} = 4\rho {l \over A}$

$= 4\left( {4\Omega } \right) = 16\Omega$