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R =  \frac{k}{A}      , where R=resistance , A=area, k=constant of variation

A thick wire would have less resistance than a thin wire.

Manipulating the equation will give us k=RA

Since A=\pi r^2

k=R\pi r^2

Given: R₁ = 5.62 Ω    r₁=2.5mm
           R₂ = ?           r₂=4.25mm

R_1\pi (r_1)^2=R_2\pi (r_2)^2

R_2 =\frac{R_1\pi (r_1)^2}{\pi (r_2)^2}

R_2 =\frac{R_1(r_1)^2}{(r_2)^2}

R_2 =\frac{(5.62)(2.5)^2}{(4.25)^2}

Have you seen the revised answer yet? Please tell me if i made another mistake. Haha
Oh man. I do have another mistake. A thicker wire would have LESS resistance than a thinner one
Really sorry about that. SInce the area is in the denominator, greater area would result to less resistance. It's also proven by the problem we just solved.
its okay haha you had the same answer as mine, soo, i think its correct and yes, its less resistance than the thin wire. Thank you again! I didn't quite understand the formulas before, but thanks to you i got to check my answer haha :)
Glad to be of help :) To commit all those mistakes, I'm probably too sleepy to be solving problems such as this haha. Good luck with your studies! ;)