Inspired by Chris Standaert

Let the perimeter be $a$ , the number of sides on the regular polygon be $n$ and the circumradius be $r$ , then:

$r\sin \left(\dfrac{\pi}{n}\right) = \dfrac{a}{2n} \implies r = \dfrac{a}{2n\sin \left(\frac{\pi}{n}\right)}$

We find that $r$ decreases with $n$ as follows:

\(\begin{array} {} n = 3 & \implies r = \dfrac{a}{6\sin \left(\frac{\pi}{3}\right)} = \dfrac{a}{3\sqrt{3}} & = 0.19245009a \\ n = 4 & \implies r = \dfrac{a}{8\sin \left(\frac{\pi}{4}\right)} = \dfrac{\sqrt{2}a}{8} & = 0.176776695a \\ n = 6 & \implies r = \dfrac{a}{12\sin \left(\frac{\pi}{6}\right)} = \dfrac{a}{6} & = 0.166666667a \\ n = 8 & \implies r = \dfrac{a}{16\sin \left(\frac{\pi}{8}\right)} & = 0.163320371a \\ n \to \infty & \implies r = \displaystyle \lim_{n \to \infty} \frac{a}{2n \sin \left(\frac{\pi}{n}\right)} = \lim_{n \to \infty} \frac{\frac{a}{2\pi}\dot{}\frac{\pi}{n}}{\sin \left(\frac{\pi}{n}\right)} = \frac{a}{2 \pi} & = 0.159154943a \end{array} \)

Therefore, the equilateral triangle has the largest circumradius.

Take a piece of string, tie it in a circle, and lay in on a flat surface. Then, using your fingers spread the string from a circle into the other shapes. The further apart you move your fingers the bigger the circumscribed circle. The equilateral triangle lets you move your fingers (three of them) apart the furthest, so this is the right shape.

One can calculate the areas and compare. However, it is true that if the area is fixed the more the "circular"( I don't know how to define it ) shape, the smaller the circumradius.

I used the fact that a circle (sphere) minimizes the surface area to volume ratio, and so extrapolated that a polygon which is least circular, would have the largest (and most inefficient) Circumradius.

The perimeter of any regular polygon is p = 2 * n * r * sin(π/n), where 'r' is the circumradius and 'n' the number of polygon sides (Re: https://proofwiki.org/wiki/Perimeter of Regular Polygon by_Circumradius). So, r = p / (2 * n * sin(π/n)) (1). The perimeter of the circle is p = 2 * n * r, so r = p / (2 * n) (2). To compare the two 'r's, we divide the right parts of the equations (1) and (2) and we need to compare 1 / π with 1 / (n*sin(π/n)) (after simplification). Now, we find that the value of n * sin(π/n) is 2.61 (min) for n=3 and 3.04 (max) for n=8. For the circle we have 3.14, which is even larger than 3.04. So, since 'r' is inversely proportional to these values, it is larger for the triangle.

This is equivalent to inscribing each figure in a unit circle and asking which has the smallest perimeter. Visualize walking the polygon perimeter as opposed to following the circle. Since a straight line is the shortest distance between two points and the triangle path is forced out to the circle the fewest times, it cuts the most distance off the circular route.

Draw a circle around each figure. The circle drawn around each polygon has a larger circumference than the polygon and therefore a larger radius than the circle. If we consider the ratio of circumference of a circle to perimeter of inscribed regular n-gon we see it is bigger for smaller n and smallest for n = 3.

Let the perimeter be $p$ .Now, we have to compare the extremes (only) as (I think) the other would follow the same pattern as all them approach near to the circle as the number of sides increase. Side of equilateral triangle(hypotenuse) = $\dfrac {p}{3}$ and the half of the other side(base) = $\dfrac{p}{6}$ . Therefore height = $\dfrac{p\sqrt3}{6}$ and the circumradius is $\dfrac {p\sqrt3}{9}$ (by the property of centroid).

Circumradius of the circle is $\dfrac {p}{2\pi}$ . Clearly $\dfrac {p\sqrt3}{9} > \dfrac{p}{2\pi}$ .

Therefore the circumradius of the triangle is greater.