It's an SAT question!

Interesting. Thanks for sharing!

To describe a rotation requires two pieces of information (See Figure 1):

a) The centre A of the rotation

b) The angle $\theta$ of the rotation

In reference to answers to the problem, the value $1.0$ means that the circle rotates $360 ^ \circ$ around the centre A . Now, if the circle rolls on a right line of a length $L$ (See Figure 2-a), the circle rotates $\theta=\dfrac{L}{2 \pi r}(360 ^ \circ)$ , where $r$ is the radius of the circle. But, what happen if a corner divides the right line? (See Figure 2-b) In this case, the circle (including the centre A ) rotates an angle $\beta<360 ^ \circ$ around the corner (See Figure 3).

So, there will not be rotation of the circle around the centre A when the circle rotates around the corner .

Finally, if the circle rolls on a hexagon (See Figure 2-c), the circle rotates $\theta=\dfrac{P}{2 \pi r}(360 ^ \circ)$ (because in the corners there are no rotation of the circle around the centre A ) where $P$ is the perimeter of the pentagon.

Therefore, if we have a regular polygon of $n$ sides, the circle rotates $\theta=\dfrac{2 R n \sin\left( \dfrac{\pi}{n} \right)}{2 \pi r}(360 ^ \circ)$ where $R$ is the circumradius of the regular polygon. When $n\rightarrow \infty$ , we have $\displaystyle\lim_{n\rightarrow \infty} n\sin\left( \dfrac{\pi}{n} \right)=\pi$ .

So $\theta=\dfrac{2 R \pi}{2 \pi r}(360 ^ \circ)=\dfrac{R}{r}(360 ^ \circ)$ , in particular, if $r=\dfrac{R}{4}$ we have $\theta=4(360 ^ \circ)$ (i.e. the value is $4$ ).