Since the sides of the square are equal and the wire is bent into a square, each side will have the same resistance, which is 20/4 or 5 Ω.
 

 Now if we make a diagonal, there will be two faces on each side of the diagonal and these two faces will have a resistance of 10 Ω.

 The resistance between the ends of a diagonal is in parallel combination.

$\therefore$${\mathrm{R}}_{\mathrm{D}}= \frac{{\mathrm{R}}_1{\mathrm{R}}_2}{{\mathrm{R}}_1+ {\mathrm{R}}_2}$

$\Rightarrow$${\mathrm{R}}_{\mathrm{D}}= \frac{10 \mathrm{X}10}{10 + 10}$

$\Rightarrow {\mathrm{R}}_{\mathrm{D}} = 5 \mathrm{\Omega }$

Therefore, the resistance between the ends of a diagonal of the square will be 5 Ω.