Answer
$I_{c} > I_{a} > I_{b}$
Work Step by Step
İf the moment of inertia at center of mass of any object is $I$, then we can calculate the moment of inertia at any point with this equation:
$I'=I+md^{2}.......................(1)$
Here, d is the distance of this point to the center of mass and m is the mass of object.
Lets assume that, after the removal of plate, the moment of inertia is $I$ at the new center of origin. So, in order to find the moment of inertia at each point, we need to calculate distance from each point to the center of mass.
Calculating location of center of mass:
Lets assume that point $ b$ is located at $(0,0)$ (left is $-x$ ,right is $+x$, above is $+y$, below is $-y$ direction). After removal of the plate, lets assume that the system consists of 3 identical plate with mass of $m$ and side length of $L$. Lets find the $x$ coordinate of center of mass:
$x_{0}=\dfrac {\sum mx}{\sum m}=\dfrac {m\times \left( -\dfrac {L}{2}\right) +m\times \left( -\dfrac {L}{2}\right) +m\times \dfrac {L}{2}}{3m}=-\dfrac {L}{6}$
$y_{0}=\dfrac {\sum my}{\sum m}=\dfrac {m\times \dfrac {L}{2}+m\times \left( -\dfrac {L}{2}\right) +m\times \dfrac {L}{2}}{3m}=\dfrac {L}{6}$
Coordinates of points and the distance from points to center of mass is:
$a\left( -L,L\right) \Rightarrow da=\sqrt {\left( y-y_{0}\right) ^{2}+\left( x-x_{0}\right) ^{2}}=\dfrac {5\sqrt {2}}{6}L\left( 2\right) $
$b\left( 0,0\right) \Rightarrow d_{b}=\sqrt {\left( y-y_{0}\right) ^{2}+\left( x-x_{0}\right) ^{2}}=\dfrac {\sqrt {2}}{6}L\left( 3\right) $
$c\left( L,-L\right) \Rightarrow d_{c}=\sqrt {\left( y-y_{0}\right) ^{2}+\left( x-x_{0}\right) ^{2}}=\dfrac {7\sqrt {2}}{6}L\left( 4\right) $
From (2),(3),(4), we get $d_{c} > d_{a} > d_{b}(5)$ and using (1) and (5), we get $I_{c} > I_{a} > I_{b}$ .
