The resistance of the piece of wire is
[tex]R= \frac{\rho L}{A} [/tex]
where
[tex]\rho = 1.68 \cdot 10^{-8}\Omega m[/tex] is the resistivity of the copper
[tex]L=5.1 cm=0.051 m[/tex] is the length of the piece of wire
[tex]A=0.13 cm^2 = 0.13 \cdot 10^{-4} m^2[/tex] is the cross sectional area of the wire
By substituting these values, we find the value of R:
[tex]R= \frac{\rho L}{A}=6.6 \cdot 10^{-5} \Omega [/tex]
Then, by using Ohm's law, we find the potential difference between the two points of the wire:
[tex]V=IR=(51 A)(6.6 \cdot 10^{-5} \Omega )=3.4 \cdot 10^{-3} V[/tex]