Answer
As long as the car moves at the minimum speed to stay on the track, the difference in apparent weight is always 6mg, and this does not depend on the radius of the loop or the actual speed of the car.
Work Step by Step
Note that the apparent weight is the normal force $F_N$ applied by the track.
Let $v_b$ be the speed at the bottom of the loop.
$F_N - mg = \frac{mv_b^2}{R}$
$F_N = \frac{mv_b^2}{R}+ mg$
We can use energy conservation to find the speed $v_t$ at the top of the loop in terms of $v_b$.
$PE_t + KE_t = KE_b$
$mgh + \frac{1}{2}mv_t^2 = \frac{1}{2}mv_b^2$
$\frac{1}{2}mv_t^2 = \frac{1}{2}mv_b^2 - 2mgR$
$v_t^2 = v_b^2 -4gR$
We can use this expression for $v_t^2$ to find $F_N$ at the top of the loop.
$F_N + mg = \frac{mv_t^2}{R}$
$F_N = \frac{mv_t^2}{R}- mg$
$F_N = \frac{mv_b^2 - 4mgR}{R}- mg$
$F_N = \frac{mv_b^2}{R}- 5mg$
At the bottom, the apparent weight is $\frac{mv_b^2}{R}+ mg$
At the top, the apparent weight is $\frac{mv_b^2}{R}- 5mg$
The difference in apparent weight at the bottom compared to the top is 6mg, which is 6 times the person's weight.
Note that as long as the car moves at the minimum speed to stay on the track, the difference in apparent weight is always 6mg, and this does not depend on the radius of the loop or the actual speed of the car.
