Solution:
First, we show that two hexagons do not suffice. Specifically, we claim that a hexagon covers less than half of the disc's boundary. First, a hexagon of side length $1$ may be inscribed in a circle, and this covers just $6$ points. Translating the hexagon vertically upward (regardless of its orientation) will cause it to no longer touch any point on the lower half of the circle, so that it now covers less than half of the boundary. By rotational symmetry, the same argument applies to translation in any other direction, proving the claim. Then, two hexagons cannot possibly cover the disc.

The disc can be covered by three hexagons as follows. Let $P$ be the center of the circle. Put three non-overlapping hexagons together at point $P$. This will cover the circle, since each hexagon will cover a $120^{\circ}$ sector of the circle.