Given $$r_{1} = 1$$ cm and $$r_{2}$$ = 6 cm Volume of 1st sphere = $$frac{4}{3} pi r_{1}^{3}$$ = $$frac{4}{3} pi 1^{3}$$ = $$frac{4}{3}pi$$ Volume of 2nd sphere = $$frac{4}{3} pi r_{2}^{3}$$ = $$frac{4}{3} pi (6)^{3}$$  Combined volume of two spheres = $$frac{4}{3} pi (217)$$  Let outer radius of hollow sphere = $$x$$ then inner radius = $$x - 1$$  Volume of the hollow sphere is given by, $$frac{4}{3} pi (x^{3} - (x - 1)^{3})$$ = $$frac{4}{3} pi (217)$$ 
$$x^{3}-(x - 1)^{3} = 217$$ 
$$x^{3}-x^{3} - 1 -3x^{2} + 3x = 217$$ 
$$3x^{2} - 3x - 216 = 0$$ 
x = 9 or -8 (which cannot be a solution) Hence, option C is the correct answer.