A uniform disc of radius R and mass M is free to oscillate about...

From the given figure, the axis of rotation is a horizontal line passing through the point $A$ on the edge of the disc. The circular arrow around the axis indicates that the disc oscillates out of its plane, meaning the axis of rotation is a tangent to the disc in its own plane.

The moment of inertia of a uniform disc about its diameter is $I_{cm} = \dfrac{1}{4}MR^2$.

Using the parallel axis theorem, the moment of inertia of the disc about the given axis of rotation is:
$I = I_{cm} + MR^2 = \dfrac{1}{4}MR^2 + MR^2 = \dfrac{5}{4}MR^2$

The distance from the center of mass of the disc to the axis of rotation is $d = R$.

The time period of a physical pendulum is given by:
$T = 2\pi\sqrt{\dfrac{I}{Mgd}}$

Substituting the values of $I$ and $d$:
$T = 2\pi\sqrt{\dfrac{\dfrac{5}{4}MR^2}{MgR}} = 2\pi\sqrt{\dfrac{5R}{4g}}$

Answer: $2\pi\sqrt{\dfrac{5R}{4g}}$