Answer
The correct banking angle of the road is $7.3^{\circ}$.
Work Step by Step
We convert the speed to units of m/s:
$v = (90~km/h)(\frac{1000~m}{1~km})(\frac{1~hr}{3600~s})$
$v = 25~m/s$
We then use the equation for the speed around a banked curve to find the banking angle;
$v = \sqrt{rg~tan(\theta)}$
$v^2 = rg~tan(\theta)$
$tan(\theta) = \frac{v^2}{rg}$
$\theta = arctan(\frac{v^2}{rg})$
$\theta = arctan(\frac{(25~m/s)^2}{(500~m)(9.80~m/s^2)})$
$\theta = 7.3^{\circ}$
The correct banking angle of the road is $7.3^{\circ}$.
