Answer
Relative to the woman, the magnitude of the acceleration at the bottom of the tire is $1500~m/s^2$
Work Step by Step
We can convert the speed of the car into units of $m/s$:
$v = (80~km/h)\times (\frac{1000~m}{1~km})\times (\frac{1~h}{3600~s}) = 22~m/s$
The velocity of the tire's center of mass is equal to the velocity of the car.
Since the car is moving at a constant velocity, the acceleration of the woman is zero, and the acceleration at the center of the tire is zero.
At the bottom of the tire, there is a centripetal acceleration directed toward the center of the tire. We can find the magnitude of this acceleration:
$a = \frac{v^2}{r}$
$a = \frac{(22~m/s)^2}{0.33~m}$
$a = 1500~m/s^2$
The magnitude of the acceleration at the bottom of the tire relative to the woman is $(1500~m/s^2-0)$ which is $1500~m/s^2$
Relative to the woman, the magnitude of the acceleration at the bottom of the tire is $1500~m/s^2$
