First find the radius of the inscribed circle of the equilateral triangle.

For an equilateral triangle with side \(a\): \(r=\dfrac{a\sqrt{3}}{6}\).

Substitute \(a=8\sqrt{6}\):

\(r=\dfrac{8\sqrt{6}\cdot\sqrt{3}}{6}=\dfrac{8\sqrt{18}}{6}=\dfrac{8\cdot3\sqrt{2}}{6}=4\sqrt{2}\text{ cm}\).

The diameter of the circle equals the diagonal of the inscribed square: \(d=2r=8\sqrt{2}\text{ cm}\).

If the side of the square is \(x\), then \(x\sqrt{2}=8\sqrt{2}\).

Therefore, \(x=8\text{ cm}\).

The radius of the circle inscribed in the square equals half the side of the square: \(R=\dfrac{6\sqrt{3}}{2}=3\sqrt{3}\text{ cm}\).

This circle is the circumcircle of the equilateral triangle.

For an equilateral triangle, \(R=\dfrac{a}{\sqrt{3}}\), so \(a=R\sqrt{3}\).

Therefore, \(a=3\sqrt{3}\cdot\sqrt{3}=9\text{ cm}\).

For a regular hexagon inscribed in a circle, the side equals the radius of the circle.

Hence, \(r=3\sqrt{3}\text{ cm}\).

This circle is inscribed in the equilateral triangle.

For an equilateral triangle, \(r=\dfrac{a\sqrt{3}}{6}\).

Thus, \(a=\dfrac{6r}{\sqrt{3}}=\dfrac{6\cdot3\sqrt{3}}{\sqrt{3}}=18\text{ cm}\).

The radius of the circle circumscribed about a square equals half the diagonal of the square.

The diagonal of the square is: \(d=12\sqrt{6}\cdot\sqrt{2}=12\sqrt{12}=24\sqrt{3}\text{ cm}\).

So the radius of the circle is: \(r=\dfrac{d}{2}=12\sqrt{3}\text{ cm}\).

This circle is inscribed in the regular hexagon, so its radius is the apothem of the hexagon.

For a regular hexagon, \(r=\dfrac{a\sqrt{3}}{2}\).

Therefore, \(a=\dfrac{2r}{\sqrt{3}}=\dfrac{2\cdot12\sqrt{3}}{\sqrt{3}}=24\text{ cm}\).

Using the circumference formula: \(2\pi R=12\pi\sqrt{3}\).

Hence, \(R=6\sqrt{3}\text{ cm}\).

For an equilateral triangle, \(R=\dfrac{a}{\sqrt{3}}\).

Therefore, \(a=R\sqrt{3}=6\sqrt{3}\cdot\sqrt{3}=18\text{ cm}\).