Masses of the three wires of same material are in the ratio of 1 : 2 : 3 and their lengths in the ratio of 3 : 2 : 1. Electrical resistance of these wires will be in the ratio of:

27 : 6 : 1

Mass, M = volume × density = $Al × d$   or   $A = M/ ld$

Resistance, R = $\rho l / A=\rho l /(M / l d)=\frac{\rho l^2 d}{M}$

So, $R \propto l^2 / M$

Thus, $R_1 : R_2 : R_3=\frac{l_1^2}{M_1}: \frac{l_2^2}{M_2}: \frac{l_3^2}{M_3}=\frac{3^2}{1}: \frac{2^2}{2}: \frac{1^2}{3}$ = 27 : 6 : 1