For a disk tangent to two sides meeting at angle $\theta,$ its center is at distance $\dfrac{r}{\sin(\frac{\theta}{2})}$ from the vertex

Use the tangency conditions to write each disk radius $r$ in terms of the polygon’s size, then compute (disk area) / (polygon area) in each case

Take the triangle with side $s.$ Three disks of radius $r$ give $s = 2r(1 + \sqrt3),$ so $T = \dfrac{3\pi r^2}{(\frac{\sqrt3}{4})s^2}$ $= \dfrac{(2\sqrt3 - 3)\pi}{2} \approx 0.73.$ For the rhombus with side $a,$ the two disks sit on the long diagonal $a\sqrt3 = 6r,$ so $r = \tfrac{a}{2\sqrt3}$ and $R = \dfrac{2\pi r^2}{(\frac{a^2\sqrt3}{2})} = \dfrac{\pi\sqrt3}{9} \approx 0.60.$ For the hexagon with side $a,$ each of the six disks touches a side at its midpoint, again giving $r = \tfrac{a}{2\sqrt3}$ and $H = \dfrac{6\pi r^2}{(\frac{3\sqrt3}{2})a^2}$ $= \dfrac{\pi\sqrt3}{9} \approx 0.60.$ So $H = R \lt T.$ Therefore, the answer is C.