The answer is $\frac{2\pi(9r_0^3-\sqrt3(r_0)^3}{13}$ First off, the radius of the first sphere is $r.$ Then, assume WLOG the radius of sphere $S_n$ is $r_n.$ Thus, we can calculate that the radius of sphere $S_{n+1} = r_{n+1}$ is simply $(r_n)/(\sqrt3).$ Proof: Since the radius of sphere $S_(n)$ is $R,$ then the side length of cube $K_(n)$ is $2(r_(n))/(\sqrt3).$ Therefore, we can compute that $r_{n+1} = (2(r_n)/(\sqrt3))/2.$ which can be simplified as $r_{n}/(\sqrt3).$ Therefore, the sum of the volume of all the spheres created in this way is just $4(r_0)^3\pi/3 + 4(r_1)^3\pi/3 + ... = 4\pi/3(((r_0)^3+((r_0)/(\sqrt3))^3+...) = \frac{2\pi(9r_0^3-\sqrt3(r_0)^3}{13}$ after applying the infinite geometric series formula. Hope this helps! Please tell me if there are any errors in my solution.