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 disc is a uniform plane lamina, so the perpendicular-axis theorem and the parallel-axis theorem can be applied.

Step 1 : Moment of inertia about a diameter
It is given that the moment of inertia of the disc about any diameter is
$$I_{\text{diameter}} = \frac{MR^{2}}{4}$$
Let two perpendicular diameters in the plane be the $$x$$- and $$y$$-axes. Hence
$$I_x = I_y = \frac{MR^{2}}{4}$$

Step 2 : Moment of inertia about the central axis normal to the disc
For a plane lamina, the perpendicular-axis theorem states
$$I_z = I_x + I_y$$
Therefore
$$I_{\text{centre, normal}} = I_z = \frac{MR^{2}}{4} + \frac{MR^{2}}{4} = \frac{MR^{2}}{2}$$

Step 3 : Shift to an axis normal to the disc through a point on its edge
The required axis is parallel to the central normal axis and is at a distance $$d = R$$ from the centre. Using the parallel-axis theorem,
$$I_{\text{edge, normal}} = I_{\text{centre, normal}} + M d^{2}$$
Substituting $$I_{\text{centre, normal}} = \frac{MR^{2}}{2}$$ and $$d = R$$:
$$I_{\text{edge, normal}} = \frac{MR^{2}}{2} + M R^{2} = \frac{3MR^{2}}{2}$$

Step 4 : Identify the value of $$x$$
The result can be written as $$\frac{x}{2}MR^{2}$$, hence
$$\frac{x}{2}MR^{2} = \frac{3}{2}MR^{2} \; \Longrightarrow \; x = 3$$

Therefore, the value of $$x$$ is $$3$$.