Stack em up!

Using the above diagram $\implies \dfrac{\dfrac{a}{2\sqrt{3}}}{R_{1}} = \dfrac{\sqrt{\dfrac{2}{3}}}{x} \implies x = 2\sqrt{2}R_{1}$

Right $\triangle{COP} \implies 9R_{1}^2 = \dfrac{2}{3}a^2 - 2\sqrt{\dfrac{2}{3}}aR_{1} + R_{1}^2 \implies$

$4R_{1}^2 + \sqrt{\dfrac{2}{3}}aR_{1} - \dfrac{1}{3}a^2 = 0 \implies$ $R_{1} = \dfrac{a}{2\sqrt{6}}$ dropping the negative root.

$H_{1} = \dfrac{2}{\sqrt{3}}$ and $R_{1} = \dfrac{a}{2\sqrt{6}} \implies H_{2} = H_{1} - 2R_{1} = \dfrac{a}{\sqrt{6}}$ and $\dfrac{H_{2}}{H_{1}} = 2 \implies$

$H_{2} = \dfrac{1}{2}H_{1} \implies R_{2} = \dfrac{1}{2}R_{1} \implies R_{3} = \dfrac{1}{2}R_{2} = \dfrac{1}{2^2}R_{1}$ and in general

$R_{n} = (\dfrac{1}{2})^{n - 1}R_{1}$

Note here that $2R_{1}\sum_{n = 1}^{\infty} R_{n} = 4R_{1} = 4(\dfrac{1}{2\sqrt{6}}) = \sqrt{\dfrac{2}{3}} = H_{1}$ .

$R_{n} = (\dfrac{1}{2})^{n - 1}R_{1} \implies V_{n} = \dfrac{4}{3}\pi (\dfrac{1}{8})^{n - 1} R_{1}^3$

$\implies V = \dfrac{4}{3}\pi R_{1}^3(\sum_{n = 1}^{\infty} (\dfrac{1}{8})^{n - 1}) =$ $\dfrac{4}{3}\pi R_{1}^3(\dfrac{8}{7}) =$

$\dfrac{4}{3}\pi (\dfrac{a}{2\sqrt{6}})^3 (\dfrac{8}{7}) = \dfrac{4}{3}\pi (\dfrac{a^3}{8 * 6\sqrt{6}})(\dfrac{8}{7}) = \dfrac{2\pi a^3}{63\sqrt{6}} = \sqrt{6}\pi \implies a^3 = 3^3 * 7 \implies$

$a =3\sqrt[3]{7} = \alpha\sqrt[3]{\beta} \implies a + b = \boxed{10}$ .