Given, Ratio of masses $=1: 3: 5$
As density will remain same
Ratio of volumes $=1: 3: 5\,\,\,\,...(i)$
and Ratio of length $5: 3: 1 \,\,\,\,...(ii)$
For ratio of resistance
$R_{1}: R_{2}: R_{3}=\frac{\rho l_1}{A_{1}}: \frac{\rho l_{2}}{A_{2}}: \frac{\rho l_{3}}{A_{3}}$
Since, resistivity remain same
$=\frac{l_{1}}{A_{1}}: \frac{l_{2}}{A_{2}}: \frac{l_{3}}{A_{3}}=\frac{l_{1}^{2}}{A_{1} l_1}: \frac{l_{2}^{2}}{A_{2} l_{2}}: \frac{l_{3}^{2}}{A_{3} l_{3}}$
From Eqs. (i) and (ii)
$R_{1}: R_{2}: R_{3}=\frac{[5]^{2}}{1}: \frac{[3]^{2}}{3}: \frac{[1]^{2}}{5} =25: 3: \frac{1}{5} $
$=125: 15: 1$