Circumscribed and Inscribed Spheres.

First we find $w$ in terms of $m$ using the inscribed sphere below.

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 $w = (\dfrac{3 + \sqrt{5}}{2})m$

Next we use the circumscribed sphere to find $\dfrac{R}{m}$ .

$R^2 = z^2 +w^2 = R^2$

$R^2 = (h - z)^2 + m^2$

$\implies z^2 + w^2 = h^2 - 2hz + z^2 + m^2 \implies z = \dfrac{h^2 - w^2 + m^2}{2h}$

Using the values for $w$ and $h$ above and simplifying we obtain:

$z = \dfrac{7 + \sqrt{5}}{4\sqrt{2}\sqrt{3 + \sqrt{5}}} m$

Using $R^2 = z^2 + w^2$ and simplifying we obtain:

$R = \dfrac{3}{4}\sqrt{\dfrac{5(7 + 3\sqrt{5})}{3 + \sqrt{5}}} m$ and $r = \sqrt{wm} = \sqrt{\dfrac{3 + \sqrt{5}}{2}} m$

$\implies \dfrac{R}{r} = \dfrac{3\sqrt{5(7 + 3\sqrt{5})}}{2\sqrt{2}(3 + \sqrt{5})} =$ $\dfrac{3}{2\sqrt{2}}\sqrt{\dfrac{5(7 + 3\sqrt{5})}{(3 + \sqrt{5})^2}} =$

$\dfrac{3}{2\sqrt{2}}\sqrt{\dfrac{5(7 + 3\sqrt{5})}{2(7 + 3\sqrt{5})}} =$ $\dfrac{3}{4}\sqrt{5} =$ $\dfrac{a}{b}\sqrt{c} \implies b * c - a = \boxed{17}$ .