This question begs for correction, as it is based on faulty reasoning.
The centripetal force is about the center of rotation, not the center of the earth. When a mass is placed in orbit, take the moon for instance, both the moon and the Earth rotate about the center of rotation, not Earths center. Both have centripetal forces about the center of rotation, and yes, they are on opposite sides of the center of rotation, and guess what...they are equal, and balance. You can do the math and prove it yourself, and that case is identical to this, the man is is not inorbit, however, his presence changes the system center of mass, and the center of rotation. He rotates about the center of rotation, not Earths center (albeit the difference is insignificantly small), and the Earth rotates about that center as well, and ghe forces do equal, and they are in fact balanced. Newtons third law (forces are in pairs) holds.
thus far in the course we have ignored that on the earths surface an object must have a slightly unbalanced force to rotate with the earth, pretending that the normal force precisely balances the objects weight. A student with a mass of 65.0kg stands at the equator. The radius of the earth is 6.38*10^6m and of course it rotates once per day.
A) what is the magnitude of the centripetal force (in newtons) required to keep the student on the earths surface?
B) Since the centripetal force is unbalanced, what is the true magnitude of the normal force on the student? (the student weighs 637 N, and with EF=ma you should find the normal force is actually less than 637 N.)
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