Answer
The road should be banked at an angle of $5.08^{\circ}$
Work Step by Step
We can assume that the frictional force is zero.
Let $F_N$ be the normal force exerted by the road on the car. We can make an equation with the vertical components of the forces acting on the car:
$\sum F_y = 0$
$F_N~cos~\theta -mg = 0$
$F_N = \frac{mg}{cos~\theta}$
The horizontal component of the normal force provides the centripetal force to keep the car moving around the curve:
$F_N~sin~\theta = m~a_r$
$(\frac{mg}{cos~\theta})~sin~\theta = m~\frac{v^2}{r}$
$g~tan~\theta = \frac{v^2}{r}$
$tan~\theta = \frac{v^2}{g~r}$
$\theta = tan^{-1}(\frac{v^2}{g~r})$
$\theta = tan^{-1}\left[\frac{(26.8~m/s)^2}{(9.80~m/s^2)(825~m)}~\right]$
$\theta = 5.08^{\circ}$
The road should be banked at an angle of $5.08^{\circ}$
