Truncated Cone.

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

$V_{T} = 2V_{s} \implies w^2 + wm + m^2 = 4wm \implies w^2 - 3wm + m^2 = 0 \implies$ $w = (\dfrac{3 \pm \sqrt{5}}{2})m$

Since $\dfrac{w}{m} > 1$ we choose $w = (\dfrac{3 + \sqrt{5}}{2})m \implies \cos(\theta) =$ $\dfrac{w - m}{w + m} = \dfrac{1 + \sqrt{5}}{5 + \sqrt{5}} = \dfrac{1}{\sqrt{5}}$

$\implies \theta \approx \boxed{63.43494882^{\circ}}$ .