on a curved path, $\tan \theta=\frac{v^{2}}{r g}$

Here, the vehicle moves with velocity

$v$ and radius of curvature $R$ and width is b.

So, here $\theta$ is small (refer figure) and thus, $\sin \theta=\frac{h}{b}=\tan \theta$

Now, we can equate $\frac{h}{b}=\frac{v^{2}}{R g}$

$\Rightarrow h=\frac{b v^{2}}{R g},$ will be the required elevation between the outer and inner edges of the road.