Consider a uniform square plate of side '$$a$$' and mass '$$m$$'. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is

For any plane lamina, the easiest way to find the moment of inertia about an axis through a point other than the centre is to start from the moment of inertia about a parallel axis through the centre of mass and then apply the Parallel-axis theorem.

1. Moment of inertia about the central axis (through centre, ⟂ to plane)
A square is a special case of a rectangle with both sides equal to $$a$$. For a rectangular lamina of sides $$l$$ and $$b$$, the moment of inertia about an axis perpendicular to its plane and passing through its centre is $$I_{\text{centre}} = \frac{m}{12}\,(l^{2}+b^{2})$$

Putting $$l=b=a$$ for the square, we get
$$I_{C} = \frac{m}{12}\,(a^{2}+a^{2}) = \frac{m}{6}\,a^{2}$$ $$-(1)$$

2. Distance between the centre and a corner
The diagonal of the square is $$a\sqrt{2}$$. Half of this diagonal is the distance from the centre to any corner: $$r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$$ $$-(2)$$

3. Apply the Parallel-axis theorem
Parallel-axis theorem: $$I = I_{C} + m r^{2}$$, where    • $$I_{C}$$ = M.I. about the central axis (from (1))    • $$r$$ = distance between the two parallel axes (from (2))

Substituting,
$$I = \frac{m}{6}\,a^{2} \;+\; m\left(\frac{a}{\sqrt{2}}\right)^{2} = \frac{m}{6}\,a^{2} \;+\; m\left(\frac{a^{2}}{2}\right) = \frac{m}{6}\,a^{2} \;+\; \frac{m}{2}\,a^{2}$$

Convert to a common denominator:
$$I = \left(\frac{1}{6} + \frac{3}{6}\right) m a^{2} = \frac{4}{6} m a^{2} = \frac{2}{3} m a^{2}$$

Thus, the moment of inertia of the uniform square plate about an axis perpendicular to its plane and passing through one of its corners is $$\boxed{\dfrac{2}{3}\,m\,a^{2}}$$

Option D which is: $$\frac{2}{3}ma^{2}$$