To solve this problem, we need to compare the moments of inertia of the sphere and the disc. For a solid sphere of radius $R$, the moment of inertia about its geometrical axis is given by:

$$I_{\text{sphere}} = \frac{2}{5} M R^2$$

where $M$ is the mass of the sphere.

When the sphere is melted into a disc of radius $r$ and thickness $t$, the volume (and hence mass) remains the same. So, the volume of the sphere is:

$$\frac{4}{3} \pi R^3$$

And the volume of the disc is:

$$\pi r^2 t$$

Equating the volumes, we get:

$$\frac{4}{3} \pi R^3 = \pi r^2 t$$

This simplifies to:

$$t = \frac{4 R^3}{3 r^2}$$

Next, we need to find the moment of inertia of the disc about the tangential axis (which is perpendicular to the plane of the disc). The moment of inertia of a disc of mass $M$ and radius $r$ about an axis through its center and perpendicular to the plane of the disc is:

$$I_{\text{disc, center}} = \frac{1}{2} M r^2$$

Using the parallel axis theorem, the moment of inertia about a tangential axis is:

$$I_{\text{disc, tangential}} = I_{\text{disc, center}} + M r^2 = \frac{1}{2} M r^2 + M r^2 = \frac{3}{2} M r^2$$

Since we are given that the moment of inertia of the sphere and the disc about their respective axes are equal, we set:

$$\frac{2}{5} M R^2 = \frac{3}{2} M r^2$$

Canceling the mass $M$ from both sides and solving for $r$, we get:

$$\frac{2}{5} R^2 = \frac{3}{2} r^2$$

Solving for $r$, we have:

$$r^2 = \frac{4}{15} R^2$$

Taking the square root of both sides:

$$r = \frac{2}{\sqrt{15}} R$$

So, the correct answer is:

Option A: $$\frac{2}{\sqrt{15}} R$$