Let's solve this step by step.

The original disc has a radius of 20 cm and is uniformly dense. Its mass (or weight) is proportional to its area:

$$M = \pi \times 20^2 = 400\pi.$$

A hole of radius 5 cm is cut out. Similarly, its mass (if it were a complete disc) would be:

$$m = \pi \times 5^2 = 25\pi.$$

Since the hole cuts through the disc, we treat it as having a negative mass. The center of the original disc is at the origin (0,0), and we need to locate the center of the hole.

The problem states that the edge of the hole touches the edge of the disc. This means that the distance between the centre of the hole and the centre of the large disc is:

$$20 \text{ cm (disc radius)} - 5 \text{ cm (hole radius)} = 15 \text{ cm}.$$

For convenience, we can assume that the hole is positioned along the positive x-axis. Thus, the centre of the hole is at:

$$\vec{r}_\text{hole} = (15, 0).$$

Now, the centre of mass (COM) of the remaining shape (the disc with the hole) can be calculated using the idea of superposition:

$$\vec{R}_{\text{cm}} = \frac{M \cdot \vec{r}_\text{disc} - m \cdot \vec{r}_\text{hole}}{M - m}.$$

Here, $\vec{r}_\text{disc} = (0,0)$ (since the large disc is centered at the origin).

Plug in the values:

$$\vec{R}_{\text{cm}} = \frac{400\pi \cdot (0,0) - 25\pi \cdot (15,0)}{400\pi - 25\pi} = \frac{-(25\pi)(15,0)}{375\pi} = \frac{-375\pi \cdot (1,0)}{375\pi} = (-1,0).$$

This tells us that the centre of mass of the remaining piece is 1 cm from the origin, along the negative x-axis.

Thus, the distance of the centre of mass from the origin is:

$$\boxed{1.0\text{ cm}}.$$

So, the correct answer is Option D.