Four point masses, each of value $$m$$, are placed at the corners of a square $$ABCD$$ of side $$\ell$$. The moment of inertia of system about an axis passing through $$A$$ and parallel to $$BD$$ is

Take the square in the XY-plane with vertices
$$A(0,0),\; B(\ell,0),\; C(\ell,\ell),\; D(0,\ell).$$

The diagonal $$BD$$ joins $$B(\ell,0)$$ to $$D(0,\ell)$$, so its slope is $$\frac{\ell-0}{0-\ell}=-1$$ and its equation is $$x+y=\ell$$.

The required axis passes through $$A(0,0)$$ and is parallel to $$BD$$. Hence its equation is
$$x+y=0.$$

For any point $$(x_0,y_0)$$ the perpendicular distance to the line $$x+y=0$$ is
$$r=\frac{|x_0+y_0|}{\sqrt{1^{2}+1^{2}}}=\frac{|x_0+y_0|}{\sqrt{2}}.$$ We now compute this distance for the three masses that are not on the axis.

Mass at $$B(\ell,0):\qquad r_B=\frac{\ell}{\sqrt{2}},\quad r_B^{2}=\frac{\ell^{2}}{2}.$$

Mass at $$C(\ell,\ell):\qquad r_C=\frac{2\ell}{\sqrt{2}}=\sqrt{2}\ell,\quad r_C^{2}=2\ell^{2}.$$

Mass at $$D(0,\ell):\qquad r_D=\frac{\ell}{\sqrt{2}},\quad r_D^{2}=\frac{\ell^{2}}{2}.$$

Moment of inertia about the axis is the sum $$I=\sum m r^{2}$$ (the mass at $$A$$ contributes zero because it lies on the axis):
$$I=m\left(\frac{\ell^{2}}{2}+2\ell^{2}+\frac{\ell^{2}}{2}\right)=m\left(3\ell^{2}\right)=3\,m\ell^{2}.$$

Option D which is: $$3\,m\ell^{2}$$