The moment of inertia of a thin rod of mass m and length L about an axis perpendicualar to it and passing through the point at a distance L/3 from the center equals

Answer 1

So here we can apply the parallel axis theorem, which relates a center of mass moment of inertia to one off-center as follows: I_new = I_center + M*(r)^2, where r is the distance moved. About the center, the moment of inertia for a perpendicular axis would have been ML^2/12. However it is a distance L/3 from the center, so I_new = I_center + M*(L/3)^2 = ML^2/12 + ML^2/9 = 7ML^2/36.

This answer is complete if the initial I_center was given to your or available in an appendix. Let me know if that was the case, otherwise I can also explain how I_center is derived from first principles.