Answer
(a) The smallest radius would be $r/4$.
(b) The smallest radius would be $r/4$.
Work Step by Step
Since static friction provides the centripetal force stopping the car from sliding, when the car is on the verge of sliding, we know $F_c=f_s^{max} (1)$
(a) From (1): $$f_s^{max}=\frac{mv^2}{r}$$
If $v$ were doubled, for $f_s^{max}$ to be maintained so that the car does not slide, $r$ would have to decrease by $2^2=4$ times. Therefore, the smallest radius would be $r/4$.
(b) We can rewrite $f_s^{max}=\mu_sF_N=\mu_smg$ (since there is no vertical acceleration), which means
$$\mu_smg=\frac{mv^2}{r}$$ $$\mu_sg=\frac{v^2}{r}$$
As you can see, $r$ in this case does not depend on the car's weight but only its speed. The smallest radius is still $r/4$.
