Volume of the wire, V= Length (l) × Area of cross-section (A)
i.e. V=lA
Let the new volume be V′ and the new area be A′
So, V′=2l×A′
Since the volume remains constant, l×A=2l×A′
⇒A′=A2
We know, R=ρlA
So, new resistance R′=ρ×2lA2
⇒ R′=4ρlA=4R=4×2=8 Ω
Hence, the resistance of the stretched wire is 8 Ω.