Nested Spheres and Regular Tetrahedrons.

The volume of the initial sphere $V_{s}(1) = \dfrac{4}{3} \pi R^3$ .

For the area of the initial tetrahedron we have:

Area of the base of the regular tetrahedron $A = \dfrac{\sqrt{3}}{4} a^2$ .

Let $AH = H$ , $BC = a$ and $BH = r^{*} \implies$

$\dfrac{a}{2} = \dfrac{\sqrt{3}}{2} r^{*} \implies r^{*} = \dfrac{a}{\sqrt{3}} \implies$ the height of the regular tetrahedron $H = \sqrt{\dfrac{2}{3}} a \implies$ the volume of the initial regular tetrahedron $V_{T}(1) = \dfrac{a^3}{6 \sqrt{2}}$ .

Finding the volume $V_{T}(1)$ of the initial tetrahedron inscribed in the initial sphere:

In right $\triangle{OBH}$ we have $OB = OA = R, OH = \sqrt{\dfrac{2}{3}} a - R$ , and $BH = \dfrac{a}{\sqrt{3}} \implies R^2 = \dfrac{a^2}{3} + ( \sqrt{\dfrac{2}{3}} a - R)^2 \implies a(a - 2 \sqrt{\dfrac{2}{3}} R) = 0, a > 0 \implies a = 2 \sqrt{\dfrac{2}{3}} R \implies V_{T}(1) = \dfrac{2}{3 \sqrt{3} \pi} V_{s}(1)$ , which is volume of the initial tetrahedron inscribed in the sphere.

Finding the volume of the inscribed sphere $V_{s}(2)$ :

Erect a perpendicular from point $O$ to a point $P$ on a face of the regular tetrahedron, then $OP = r = OH, AP = \dfrac{a}{\sqrt{3}},$ and $OA = \sqrt{\dfrac{2}{3}} a - r$ .

In right $\triangle{AOP}$ we have: $R^2 + \dfrac{a^2}{3} = (\sqrt{\dfrac{2}{3}} a - r)^2 \implies a(\dfrac{a}{3} - 2\sqrt{\dfrac{2}{3}} r) = 0, a> 0 \implies$ $r = \dfrac{a}{2\sqrt{6}}$ and $a = 2\sqrt{\dfrac{2}{3}} R \implies r = \dfrac{R}{3} \implies V_{s}(2) = \dfrac{1}{27} V_{s}(1).$

Let $a_{1} = a = 2\sqrt{\dfrac{2}{3}} R, r_{1} = R$ and $r_{2} = r = \dfrac{R}{3}$

$\implies a_{2} = 2(\sqrt{\dfrac{2}{3}}) r_{2} = 2(\sqrt{\dfrac{2}{3}}) (\dfrac{1}{3}) R, r^3 = \dfrac{r_{2}}{3} = \dfrac{R}{3^2}, a_{3} = 2(\sqrt{\dfrac{2}{3}}) r_{3} = 2(\sqrt{\dfrac{2}{3}}) (\dfrac{1}{3})^2 R$

In General:

$r_{n} = (\dfrac{1}{3})^{n - 1} R$ and $a_{n} = 2(\sqrt{\dfrac{2}{3}}) (\dfrac{1}{3})^{n - 1} R \implies V_{s}(n) = (\dfrac{1}{27})^{n - 1} V_{s}(1)$ and $V_{T}(n) = \dfrac{2}{3\sqrt{3} \pi} (\dfrac{1}{27})^{n - 1} V_{s}(1) \implies$

$V_{s} = V_{s}(1) * \sum_{n = 0}^{\infty} (\dfrac{1}{27})^{n} = \dfrac{27}{26} V_{s}(1)$ and $V_{T} = \dfrac{2}{3\sqrt{3} \pi} V_{s}(1) * \sum_{n = 0}^{\infty} (\dfrac{1}{27})^{n} = \dfrac{2}{3\sqrt{3} \pi} (\dfrac{27}{26}) V_{s}(1) \implies$ $V_{s} * V_{T} = \dfrac{3^4 * \sqrt{3}}{2 * 13^2 \pi} (V_{s}(1))^2 = \dfrac{a^4 \sqrt{a}}{bc^2 \pi} (V_{s}(1))^2$ $\implies a + b + c = \boxed{18}.$