= Solution
Measure position $x$ along the ideal thin rod from its <centre of mass>. Its perpendicular distance from the inclined axis is $|x|\sin\alpha$, and its <linear mass density> is $M/L$. The <moment of inertia> is therefore
$$
I=\frac ML\int_{-L/2}^{L/2}x^2\sin^2\alpha\,dx
=\boxed{\frac{ML^2}{12}\sin^2\alpha.}
$$
This tends to zero for an axis along an infinitely thin rod and to $ML^2/12$ for a perpendicular central axis. A finite rod thickness would add a contribution absent from the stated thin-rod model.
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