From a square sheet of uniform density, one of the four portions within its diagonals is removed .. Find the of centre of mass of a remaining portion if the side of the square is a. The portion removed is right one

Lets find out the com of the triangle first.
Let the density be $\rho$
The center of the squre is taken as the origin
The x coordinate of the CoM is
$x_{cm}=\frac{{\int }_0^{a/2}2\rho xx\tan \theta dx}{\rho {\displaystyle \frac{a^2}{4}}}=\frac{8}{a^2}\tan \left({45}^0\right){\int }_0^{a/2}{x}^{2}dx=\frac{8}{a^2}\frac{{\left({\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^3}{3}=\frac{a}{3}$
Now using symmetry $y_{cm}=0$
Similarly, the coordinates of the other three triangles are (-a/3,0), (0,a/3),(0,-a/3)
So the CoM of the remaining square is
$$\begin{gathered} (X_{cm},Y_{cm})=\left(\frac{-{\displaystyle \frac{a}{3}}+0+0}{3},\frac{0+{\displaystyle \frac{a}{3}}-{\displaystyle \frac{a}{3}}}{3}\right) \\ \Rightarrow (X_{cm},Y_{cm})=\left(-\frac{a}{9},0\right) \end{gathered}$$