$\because$ 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) $
$=25 \times 5=125 \Omega$
