Moment of inertia of a square plate of side $$l$$ about the axis passing through one of the corner and perpendicular to the plane of square plate is given by:

We are asked to find the moment of inertia of a uniform square plate of side $$l$$ about an axis which (i) passes through one of its corners and (ii) is perpendicular to the plane of the plate. To proceed, we shall first recall the standard result for a simpler, symmetrical axis and then use the Parallel-Axis Theorem to shift that result to the required corner axis.

For a thin rectangular plate of mass $$M$$, sides $$a$$ and $$b$$, the moment of inertia about an axis through its centre and perpendicular to its plane is given by the well-known formula

$$I_{\text{CM}}=\frac{M}{12}\left(a^{2}+b^{2}\right).$$

In the present problem the plate is a square, so $$a=b=l$$. Substituting $$a=l$$ and $$b=l$$ into the above formula we obtain

$$I_{\text{CM}}=\frac{M}{12}\left(l^{2}+l^{2}\right)=\frac{M}{12}\,2l^{2}=\frac{M l^{2}}{6}.$$

Thus the moment of inertia about the central (centroidal) axis, which is perpendicular to the plane, equals $$\displaystyle\frac{M l^{2}}{6}$$.

However, the axis demanded in the question is not through the centre but through a corner. To relate the two axes we invoke the Parallel-Axis Theorem. The theorem states:

$$I_{\text{new}} \;=\; I_{\text{CM}} + M d^{2},$$

where $$d$$ is the distance between the centre of mass and the new axis. In our square, the centre lies at the midpoint of both diagonals, i.e.\ at coordinates $$\left(\tfrac{l}{2}, \tfrac{l}{2}\right)$$ when one corner is taken as the origin. The distance from this centre to that same corner is therefore

$$d=\sqrt{\left(\frac{l}{2}\right)^{2}+\left(\frac{l}{2}\right)^{2}}=\sqrt{\frac{l^{2}}{4}+\frac{l^{2}}{4}}=\sqrt{\frac{l^{2}}{2}}=\frac{l}{\sqrt{2}}.$$

We next square this distance, because the parallel-axis term uses $$d^{2}$$:

$$d^{2}=\left(\frac{l}{\sqrt{2}}\right)^{2}=\frac{l^{2}}{2}.$$

Now we substitute $$I_{\text{CM}}=\dfrac{M l^{2}}{6}$$ and $$d^{2}=\dfrac{l^{2}}{2}$$ into the Parallel-Axis Theorem:

$$I_{\text{corner}} \;=\; I_{\text{CM}} + M d^{2} =\frac{M l^{2}}{6} + M\left(\frac{l^{2}}{2}\right).$$

We combine the two fractions by expressing both over a common denominator of 6:

$$I_{\text{corner}} =\frac{M l^{2}}{6} + \frac{3 M l^{2}}{6} =\frac{4 M l^{2}}{6}.$$

Finally, we reduce $$\dfrac{4}{6}$$ to its simplest form $$\dfrac{2}{3}$$, giving

$$I_{\text{corner}}=\frac{2}{3}M l^{2}.$$

This value exactly matches Option B in the list provided. Hence, the correct answer is Option B.