Answer
The normal force on the car when it is at the bottom of the track is 21.7 N.
Work Step by Step
At the top of the track:
$\sum F = \frac{mv^2}{r}$
$F_N+mg = \frac{mv^2}{r}$
$F_N = \frac{mv^2}{r}-mg$
At the bottom of the track:
$\sum F = \frac{mv^2}{r}$
$F_N-mg = \frac{mv^2}{r}$
$F_N = \frac{mv^2}{r}+mg$
The normal force at the bottom of the track is $2mg$ greater than the normal force at the top of the track.
$F_N = 6.00~N + 2mg$
$F_N = 6.00~N+(2)(0.800~kg)(9.80~m/s^2)$
$F_N = 21.7~N$
The normal force on the car when it is at the bottom of the track is 21.7 N.
