Radius of curved road is ' R ', width of road is ' b '. The outer edge of road is raised by ' $h$ ' with respect to inner edge so that a car with velocity ' $V$ ' can pass safe over it, then value of ' $h$ ' is ( $g=$ acceleration due to gravity)

For a car moving with velocity $V$ on a banked road of radius $R$, the horizontal component of the normal force provides the centripripetal force, while the vertical component balances gravity. For a banking angle $\theta$:

$N \sin\theta = \frac{mV^2}{R}$
$N \cos\theta = mg$

Dividing these expressions yields $\tan\theta = \frac{V^2}{Rg}$.

From the road geometry with width $b$ and outer edge raised by $h$, $\tan\theta = \frac{h}{b}$. Equating gives:

$\frac{h}{b} = \frac{V^2}{Rg}$

Solving for the height difference: $h = \frac{V^2 b}{Rg}$, which corresponds to option A.