I've seen it before - 2

We know $A_{0}A_{1}$ is 1 so we only need to find $A_{0}A_{2}$ and $A_{0}A_{4}$ Alt text

We call $O$ the center and $M$ the intersection of $A_{0}A_{2}$ and $O A_{1}$ . A right angled triangle is formed by $A_{0}M$ and $M A_{1}$ , so by Pythagorean Theorem we can calculate the length of $A_{0} M$ :

$1-.5^2=.75$

So the complete length of $A_{0}A_{2}$ its $2\sqrt{.75}$ that equals to $\sqrt{2^2\times .75}$ $=$ $\sqrt{3}$ . Since $A_{0}A_{2}$ $=$ $A_{0}A_{4}$ ,our desired number is:

$\sqrt{3}\times\sqrt{3}$ $=$ $\boxed{3}$

Clearly, each side length of this regular hexagon is unity $\Rightarrow A_{0}A_{1} = 1$ . For the remaining two lengths, let us deploy the Law of Cosines to congruent triangles $\triangle{A_{0}A_{1}A_{2}}$ and $\triangle{A_{0}A_{4}A_{5}}:$

$(A_{0}A_{2})^2 = (A_{0}A_{4})^2 = (A_{0}A_{1})^2 + (A_{1}A_{2})^2 - 2 \cdot (A_{0}A_{1})(A_{1}A_{2}) \cos(2\pi/3) = 2(1^2) - 2(1^2)(-\frac{1}{2}) =3;$

or $A_{0}A_{2} = A_{0}A_{4} = \sqrt{3}.$ Thus, $A_{0}A_{1} \cdot A_{0}A_{2} \cdot A_{0}A_{4} = 1 \cdot (\sqrt{3})^2 = \boxed{3}.$

Hexagon has six sides..given unit radius..so the length is 6...we then can divide by 2 for each line segment...A0,A1=6/2=3