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

Given the diameter $ r $ of the ring, the radius $ R $ is $ \frac{r}{2} $.

The formula for the moment of inertia about a tangential axis in the plane of the ring is:

$ I_{\text{tangential}} = \frac{3}{2} M \left(\frac{R}{2}\right)^2 $

Substitute the value of $ R = \frac{r}{2} $ into the equation:

$ I_{\text{tangential}} = \frac{3}{2} M \left(\frac{r}{2}\right)^2 = \frac{3}{8} Mr^2 $

Therefore, the moment of inertia of the ring about the specified axis is $\frac{3}{8} Mr^2$.