Answer
$I = \frac{1}{12}ML^2+\frac{1}{4}M_1~L^2+\frac{1}{16}M_2~L^2$
Work Step by Step
To find the moment of inertia of the object, we can add the moment of inertia of the rod, the moment of inertia of $M_1$, and the moment of inertia of $M_2$.
$I =\frac{1}{12}ML^2+M_1(\frac{L}{2})^2+M_2(\frac{L}{4})^2$
$I = \frac{1}{12}ML^2+\frac{1}{4}M_1~L^2+\frac{1}{16}M_2~L^2$