when car moves along a circular path then centripetal force for taking turn will be provided by the component of normal force
Let the road is tilted by some angle
so force equations are given as
[tex]F_n cos\theta = mg[/tex]
[tex]F_n sin\theta = \frac{mv^2}{R}[/tex]
now divide two equations
[tex]tan\theta = \frac{v^2}{Rg}[/tex]
now as it is given to us that
[tex]v = 80 km/h[/tex]
[tex]v = 80* \frac{5}{18} = 22.2 m/s[/tex]
[tex]R = 50 m[/tex]
now we have
[tex]tan\theta = \frac{22.2^2}{50*9.8}[/tex]
[tex]tan\theta = 1[/tex]
[tex]\theta = tan^{-1}1[/tex]
[tex]\theta = 45^0[/tex]
so it is inclined at an angle of 45 degree