The electrical resistance of the wire is proportional to its length L and inversely proportional to the square of its diameter, [tex]1/r^2[/tex].
The length of the wire in the problem is changed from 200 inches to 500 inches, so the new length is 2.5 times the initial length:
[tex] \frac{L_1}{L_0}= \frac{500}{200}=2.5 [/tex]
the diameter of the wire is not changed, so the new electrical resistance must be 2.5 times the original value:
[tex]R_1 = 2.5 R_0 = 2.5 \cdot 20 \Omega =50 \Omega[/tex]
