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A car of weight W = 11.5 kN makes a turn on a track that is banked at an angle of θ = 22.3°. Inside the car, hanging from a short string tied to the rear-view mirror, is an ornament. As the car turns, the ornament swings out at an angle of φ = 29.5° measured from the vertical inside the car. What is the force of static friction between the car and the road

Hello Minaya. I suggest you set this up by first drawing a free body diagram the ornament: it hanges from the vertical at an angle of $\theta + \phi$ . The forces acting on it are tension in the string, gravity, and centripedal acceleration. Draw the diagram, and you should be able to wrirte an expression for $\omega^2R$ in terms of the angles $\theta$ and $\phi$ .

Now consider the car - the friction force that keeps it from sliding must equal the centripedal force in the direction of the road surface, which is the car's mass $M \times \omega^2 R \times \cos(\theta)$ . You should get a result that involves only M, g and the two angles you were given.