Answer
$R_{elongation} = 4R_{initial}$
Work Step by Step
Recall that $R = \displaystyle \frac{\rho L}{A}$, where $\rho$ is the resistivity of the material the wire is made from.
Since in this example, the material is not changing, neither is the resistivity. So $R \propto \displaystyle \frac{L}{A}$
The trick to this problem is that the volume stays constant over elongation, that is, doubling the length of the wire will not change the volume. This means that the cross-sectional area will be changed by a factor as well. We can find that factor by:
$V_{initial} = V_{final} \quad\rightarrow\quad A_{initial}L_{initial} = A_{final}L_{final}$
and we know that $2L_{initial} = L_{final}$
$A_{initial}L_{initial} = A_{final}(2L_{initial})$
$A_{initial} = 2A_{final}$
$\displaystyle \frac{A_{initial}}{2} = A_{final}$
We can now find the resistance of the elongated wire:
$R_{elongation} = \displaystyle \frac{\rho L_{final}}{A_{final}} = \frac{\rho (2L_{initial})}{A_{initial}/2} = 4R_{initial}$
