Since rate of heat $$ = {{{V^2}} \over R} \Rightarrow $$ rate of heat $$ \propto {1 \over R}$$, where $$R = {{\rho L} \over A} = {{\rho L} \over {\pi {r^2}}}$$; here r is radius of wire and L is length of wire. Therefore,

$$R \propto {L \over {{r^2}}}$$.

Thus, rate of heat $$ \propto {{{r^2}} \over L}$$.

When length in halved and radius is doubled then, rate of heat $$ \propto {{{{(2r)}^2}} \over {L/2}} = {{8{r^2}} \over L}$$

Therefore, rate of heat developed in wire will be increased 8 times.