Moment of inertia of a disc of mass $$M$$ and radius '$$R$$' about any of its diameter is $$\frac{MR^2}{4}$$. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, $$\frac{x}{2}MR^2$$. The value of $$x$$ is ______.

The moment of inertia of a disc of mass $$M$$ and radius $$R$$ about any of its diameters is $$\frac{MR^2}{4}$$. We seek the moment of inertia about an axis perpendicular to the disc and passing through a point on its edge.

By the perpendicular axis theorem for a planar body, $$I_z = I_x + I_y$$. Since each diameter contributes $$I_x = I_y = \frac{MR^2}{4}$$, the moment of inertia about the central axis normal to the plane is $$I_{\text{center}} = \frac{MR^2}{4} + \frac{MR^2}{4} = \frac{MR^2}{2}$$.

Applying the parallel axis theorem to shift the axis to the edge, which is at a distance $$R$$ from the center, gives $$I_{\text{edge}} = I_{\text{center}} + MR^2 = \frac{MR^2}{2} + MR^2 = \frac{3MR^2}{2}$$.

Since the general form is $$I_{\text{edge}} = \frac{x}{2}MR^2$$, equating to $$\frac{3MR^2}{2}$$ yields $$\therefore x = \boxed{3}$$.