As shown above we know that the $L_{bar} = .864 m$ and that $M_{bar} = .833 kg$
Using this we can easily find the rotational inertia of a bar by using the formula $I_{bar} = \frac{1}{12}ML^2$ Which plugged in with the given numbers gives roughly $51.8 * 10^{-3}$, which ended up being correct.
For the cylinder masses you simply use the general intertia equation for a point mass, $I = MR^2$
Where $M_{cyl} = 2.65 kg$ and $R_{cyl} = .2 m$ which when plugged into the equation gives $106 * 10^{-3}$ which was also correct.
So to find the total of the apparatus I was under the impression that you just sum up the inertia from the bar and the two cylinders or in other words $I_{tot} = 51.8 + 106(2)$ which gives $263.8$ which is apparently false! So, what am I doing wrong? Any pointers? Thanks!