Answer
$I_{b} > I_{c} > I_{a}$
Work Step by Step
Moment of inertia of cylinders about their central axes is:
$I=\dfrac {mR^{2}}{2}$
Given the values, the moment of inertia is calculated as:
$I_{a}=\dfrac {m_{1}R^{2}_{1}}{2}=\dfrac {26kg\times 1m^{2}}{2}=13kgm^{2}\left( 1\right) $
$I_{b}=\dfrac {m_{2}R^{2}_{2}}{2}=\dfrac {7kg\times\left( 2m\right) ^{2}}{2}=14kgm^{2}(2)$
$I_{c}=\dfrac {m_{3}R^{2}_{3}}{2}=\dfrac {3kg\times\left( 3m\right) ^{2}}{2}=13,5kgm^{2}\left( 3\right) $
So from (1),(2) and (3) we get
$I_{b} > I_{c} > I_{a}$
