A metallic wire has resistance R. Some part of it is stretched so that its   length becomes doubled. Now the resistance of the  total wire becomes$\frac{4R}{3}$ . The ratio of the lengths of unstretched part  and stretched part of initial wire is 

suppose the resistance of the stretched part is $R_1$ and that of unstretched  part is $R_2$.   then $R_1+R_2 =R$------(1)

after stretching $4 R_1+R_2 =\frac{4R}{3}$  -----(2)       $\left(if length is doubled resis\tan ce becomes 4 times\right)$

solving we get $3R_1=\frac{R}{3}$  $R_1=\frac{R}{9}$  then $R_2=\frac{8}{9}$

$\frac{R_2}{R_1}=\frac{8}{1}$