Verified answer

In order to find average (by time) distance you also need to know eccentricity of orbit.

Even then, the problem if far from elementary.

The average inverse distance is far easier, because it co-insides with average potential energy, and can be found by virial.

Your teacher probably wants major semi-axis.

It's length is ³√(410/365)² = 1.08 a.u.

The force of gravity from the Sun = F = GmM/R^2; where M is Sun mass, R is the distance between the Sun center and the center of a planet of mass m, and G is a constant.

Centripetal force C = mv^2/R = GmM/R^2 = F when the planet is in orbit with tangential velocity v = WR; where W = angular velocity = 1/T and T is the period of revolution around the Sun.

Solve for R, which is the mean distance from the Sun. v^2 = GM/R; so that, R = GM/v^2 = GM/(WR)^2 and R^3 = GM/W^2 = GMT^2. As G and M are constants, we can write R^3 = kT^2. Now let's call R the radius of Ikkie's orbit around the Sun and r the radius of Earth's orbit around the Sun. T = 410 days for Ikkie and t = 365 days for Earth, their respective periods for one revolution each.

Now set the ratio R/r = k^1/3 T^2/3//k^1/3 t^2/3; so that R = r(T/t)^2/3 since the k factors cancel out; and there you have the mean distance of Ikkie (R) from the Sun. r = 93 million miles or 1 AU. T = 410 days and t = 365 days. You can do the math.

Lesson, use ratios when comparing similar physical characteristics (like orbital distances R and r) that depend on the same factors (like G, m, v, and M). Often, as in this case, most of those other factors will cancel out in the ratio, leaving only the factors that matter (like T).

By the way, we just derived a version of Kepler's Law on planetary orbits. Look it up, you'll see what I mean. Also, some of the answers alluding to elliptical paths, rather than the circular one I assumed, are correct. But as none of the eccentricity information was provided, I assumed circular orbits for both Earth and Ikkie.

The orbit varies. Here's a link.