Answer
The inner diameter of the sphere is $~~57.3~cm$
Work Step by Step
The weight of the water displaced by the sphere is equal to the weight of the sphere.
Then the mass of the water displaced by the sphere is equal to the mass of the sphere.
We can find the mass of the displaced water:
$m = \rho~V$
$m = \rho~\frac{4}{3}\pi~r_2^3$
$m = (1000~kg/m^3)~(\frac{4}{3})(\pi)~(0.300~m)^3$
$m = 113.1~kg$
Then the mass of the sphere is $113.1~kg$
We can find the volume of the iron used to make the sphere:
$V = \frac{m}{\rho}$
$V = \frac{113.1~kg}{7870~kg/m^3}$
$V = 0.0144~m^3$
We can find $r_1$, the inner radius:
$V = \frac{4}{3}\pi~r_2^3-\frac{4}{3}\pi~r_1^3$
$\frac{4}{3}\pi~r_1^3 = \frac{4}{3}\pi~r_2^3 - V$
$r_1^3 = r_2^3 - \frac{3V}{4\pi}$
$r_1 = \sqrt[3] {r_2^3 - \frac{3V}{4\pi}}$
$r_1 = \sqrt[3] {(0.300~m)^3 - \frac{(3)(0.0144~m^3)}{4\pi}}$
$r_1 = 0.2867~m$
$r_1 = 28.67~cm$
Then the inner diameter of the sphere is $~~57.3~cm$