The distance between its perihelion and aphelion is 10A.U. What is its orbital period in term of years?

Was this a homework question? Are you sure it said "perihelion and aphelion of the sun"? Or was it perihelion and aphelion of a hypothetical planet?

The distance between perihelion and aphelion is called the major axis of the ellipse, and half that distance is called the "semi-major axis", in this case 5 AU. That's usually used as the "average distance from the sun".

This sounds like a question about Kepler's Third Law, which says that for objects orbiting the sun, T^2 = R^3 where R = semi major axis in AU and T = period in years. So plug in 5 for R and solve for T.

Other answers are incorrect. We do indeed know the period if all we know is the semi-major axis. That's what Kepler tells us, and Kepler's Laws are derived directly from Newton's Law of Gravitation.

Impossible to tell. The object could be in an orbit where 10 AU means that the orbit is almost circular, and the period is measured in millions of years.

Or you could be talking about a nearby comet, that takes about 30 years to orbit.

The Sun does not have a perihelion: it is never any distance away from itself. You can't use this kind of information to calculate an orbital period even for the planets. You need the semi-major axis, which in laymans terms is the average distance between the satellite and its parent body. You can't derive that from the difference between aphelion and perihelion.

There are an infinite number of possible orbits for which the perihelion is 10AU less than the aphelion. There is no way to determine the orbital period without knowing the perihelion or aphelion distances.

If it is orbiting the Sun (which I infer from "perihelion" instead of "periapsis" or "periastron") Randy P.'s answer will do the trick.

The the combined mass of the object and what it is orbiting is significantly different than 1 solar mass, you need a more complex version of Kepler's Third Law.