$\because$ Volume of wire is constant in stretching

$V _{ i }= V _{ f }$

$A _{ i } \ell_{ i }= A _{ f } \ell_{ f }$

$A \ell= A ^{\prime}(5 \ell)$

$A ^{\prime}=\frac{ A }{5}$

$R _{ f }=\frac{\rho \ell_{ f }}{ A _{ f }}=\frac{\rho(5 \ell)}{\left(\frac{ A }{5}\right)}$

$=25\left(\frac{\rho \ell}{A}\right)$

$=25 \times 5=125 \Omega$