Answer
$AB=DE\lt CD\lt BC$
Work Step by Step
In sections AB and DE, the car moves straight at a constant speed, so there is no acceleration in these two sections.
In sections BC and CD, there is centripetal acceleration. We know that $$a_c=\frac{v^2}{r}$$
The car maintains its speed, so $v$ is constant. Radius of turning circle in BC is smaller than that in CD, so $a_c$ in BC is larger than $a_c$ in CD.
So, ranking the accelerations, we end up with $AB=DE\lt CD\lt BC$
