Answer
(a) It decreses by a factor of two.
Work Step by Step
Let $R_i $ be the initial resistance and $R_f $ be the final resistance.
$ R_i = \frac{\rho *l}{A}$
When length and diameter are doubled
$ l' = 2l \\ d' =2d \implies r' =2r$
$ \therefore A' = \pi r'^2 = \pi (2r)^2 = 4 \pi r^2 = 4A$
Thus $R_f =\frac{ \rho l'}{A'} =\frac{ \rho *2l}{4A} =\frac{1}{2} R_i $
hence resistance decreases by a factor of two.
