Truncated Square Pyramids

Using the fact that tangents to a circle from an outside point are congruent we obtain the diagram above.

The height $h$ of the truncated square pyramid is $h = 2r$

Using right $\triangle{ABC}$ we have:

$(w + m)^2 = (w - m)^2 + 4r^2 \implies w^2 + 2wm + m^2 = w^2 - 2wm + m^2 + 4r^2 \implies 4wm = 4r^2 \implies r^2 = wm \implies$

$r = \sqrt{wm}$ .

The volume of the inscribed sphere $V_{s} = \dfrac{4}{3}\pi(wm)^{\frac{3}{2}}$

and

The volume of the truncated cone is $V_{T} = \dfrac{2\pi}{3}(4w^2 + 4wm + 4m^2)(wm)^{\frac{1}{2}}$

$= \dfrac{8\pi}{3}(w^2 + wm + m^2)(wm)^{\frac{1}{2}}$

$V_{T} = 2V_{s} \implies w^2 + wm + m^2 = \pi wm \implies w^2 - (\pi - 1)wm + m^2 = 0 \implies$ $w = (\dfrac{\pi - 1 \pm \sqrt{(\pi - 1)^2 - 4}}{2})m$

Since $\dfrac{w}{m} > 1$ we choose $w = (\dfrac{\pi - 1 + \sqrt{\pi^2 - 2\pi - 3}}{2})m$

and $h = 2\sqrt{wm} = (2\sqrt{\dfrac{\pi - 1 + \sqrt{\pi^2 - 2\pi - 3}}{2}})m$

$\implies \tan(\theta) = \dfrac{h}{w} = \dfrac{2\sqrt{2}}{\sqrt{\pi - 1 + \sqrt{\pi^2 - 2\pi - 3}}}$

$\implies \theta \approx \boxed{58.91655379^{\circ}}$