Three identical cars, A , B , and C , travel at the same speed...

For car $A$ moving on a flat bridge, the vertical acceleration is zero. The normal force is given by $F_A = mg$.

For car $B$ moving on a convex upward bridge, the centripetal acceleration is directed downwards towards the center of curvature. The equation of motion is $mg - F_B = \dfrac{mv^2}{R}$, which gives $F_B = mg - \dfrac{mv^2}{R}$.

For car $C$ moving on a concave upward bridge, the centripetal acceleration is directed upwards towards the center of curvature. The equation of motion is $F_C - mg = \dfrac{mv^2}{R}$, which gives $F_C = mg + \dfrac{mv^2}{R}$.

Comparing the three normal forces, we get $F_C > F_A > F_B$.

Therefore, $F_C$ is the largest of the three forces.