The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is :

The diameter of the ring is given as $$r$$, hence its radius is
$$R = \frac{r}{2}\,.$$

Step 1: Moment of inertia about a diameter (axis in the plane, through centre)
Every mass element of a thin ring is at the same distance $$R$$ from the centre. For an axis along a diameter (say the $$x$$-axis) lying in the plane of the ring, the perpendicular distance of a point $$\bigl(R\cos\theta,\;R\sin\theta\bigr)$$ from that axis is $$R\lvert\sin\theta\rvert$$. Using $$I = \int r_\perp^{\,2}\,dm$$,

$$I_{\text{diameter}} = \int_0^{2\pi} \! \left(R\sin\theta\right)^2 \left(\frac{M}{2\pi}\,d\theta\right) = \frac{MR^2}{2}\,. $$

Step 2: Moment of inertia about a tangential axis in the plane
The required axis is parallel to the diameter axis but touches the ring at one point, so it is shifted by a distance $$R$$ from the centre. Apply the Parallel-Axis Theorem:

$$I_{\text{tangent}} \;=\; I_{\text{diameter}} + M R^2 = \frac{MR^2}{2} + M R^2 = \frac{3}{2}\,M R^2 \,.$$

Step 3: Express in terms of the given diameter
Substitute $$R = \dfrac{r}{2}$$:

$$I_{\text{tangent}} = \frac{3}{2}\,M\left(\frac{r}{2}\right)^2 = \frac{3}{2}\,M\,\frac{r^{2}}{4} = \frac{3}{8}\,M r^{2}\,.$$

Therefore, the moment of inertia of the circular ring about a tangential axis lying in its own plane is $$\frac{3}{8}Mr^{2}$$.

The correct choice is Option B.