Answer
a) $ {\rm(5.71\;cm, 4.57\;cm)}$
b) $ 1.71\times 10^{-3}\;\rm kg\cdot m^2 $
Work Step by Step
a) To find the coordinates of the center of the mass of the given object, we need to find $X_{cm}$ and $Y_{cm}$.
We choose point A to be our origin at which $(x=0,y=0)$
$$X_{cm}=\dfrac{m_Ax_A+m_Bx_B+m_Cx_C+m_Dx_D}{m_A +m_B +m_C +m_D }$$
Plugging the known;
$$X_{cm}=\dfrac{ 100 (0) +200(0)+200(10)+200(10)}{100+200+200+200}=\color{red}{\bf 5.71}\;\rm cm$$
And
$$Y_{cm}=\dfrac{m_Ay_A+m_By_B+m_Cy_C+m_Dy_D}{m_A +m_B +m_C +m_D }$$
Plugging the known;
$$Y_{cm}=\dfrac{ 100 (0) +200(8)+200(8)+200(0)}{100+200+200+200}=\color{red}{\bf 4.57}\;\rm cm$$
Therefore, the coordinates of the center of mass of the four masses are
$$ {\bf (5.71\;cm, 4.57\;cm)}$$
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b)
The moment of inertia around $\overline{BD}$ is given by
$$I_{BD}=m_Ar_A^2+m_Cr_C^2 \tag 1$$
Now we need to find $r_A$ and $r_C$.
From the geometry of the first figure below,
$$\tan\theta=\dfrac{8}{10}$$
Hence,
$$\theta=\tan^{-1}\left( \dfrac{8}{10} \right)=\bf 38.66^\circ $$
Now, from the second figure below, the yellow right triangle.
$$\sin\theta=\sin38.66^\circ=\dfrac{r_C}{10}$$
Hence,
$$r_C=10\sin38.66^\circ\tag 2$$
From the violet right triangle below,
$$\sin\theta=\sin38.66^\circ=\dfrac{r_A}{10}$$
Hence,
$$r_C=10\sin38.66^\circ\tag 3$$
We can see that $r_A=r_C$
Plugging (2) and (3) into (1); and converting the units to SI units.
$$I_{BD}=m_A(0.10\sin38.66^\circ)^2+m_C(0.10\sin38.66^\circ)^2 $$
Plugging the known;
$$I_{BD}=(0.100)(0.10\sin38.66^\circ)^2+(0.200)(0.10\sin38.66^\circ)^2 $$
$$I_{BD}=\color{red}{\bf 1.71\times 10^{-3}}\;\rm kg\cdot m^2 $$
