The resistance of the cylindrical wire is [tex] R=\frac{\rho l}{A} [/tex].
Here [tex] R [/tex] is the resistance, [tex] l [/tex] is the length of the wire and [tex] A [/tex] is the area of cross section. Since the wire is cylindrical [tex] A=\frac{\pi d^2}{4} [/tex]. Rearranging the above equation,
[tex] A=\frac{\rho l}{R}\\
\frac{\pi d^2}{4}=\frac{\rho l}{R}\\
d=\sqrt{\frac{4\rho l}{\pi R}}
[/tex]
Here [tex] l=120, R=6, \rho=1.68(10^{-8}) [/tex].
Substituting numerical values,
[tex] d=\sqrt{\frac{4(1.68)(10^{-8}) (120)}{\pi (6)}}\\
d=0.0006541 [/tex]
Te diameter of the wire is [tex] 0.6541 mm [/tex]
