When a wire is stretched, its volume remains constant. Volume $V = AL = \pi r^2 L$. If the radius becomes $r/2$, the new area is $A' = \pi(r/2)^2 = A/4$. Since volume is constant, the new length $L' = 4L$. Resistance $R = \rho L/A$. The new resistance $R' = \rho L'/A' = \rho(4L)/(A/4) = 16 (\rho L/A) = 16R$.