Answer
$R=1.8 \times 10^5 \mathrm{~m} $
$\frac{V_s}{V_E}=2.3 \times 10^{-5}$
Work Step by Step
(a) The mass of an iron atom is
$$
m=56 \mathrm{u}=\mathrm{D}_{56 \mathrm{u}}\left[\mathrm{f} .66 \times 10^{-27} \mathrm{~kg} / \mathrm{u} h\\=9.30 \times 10^{-26} \mathrm{~kg} .\right.
$$
Therefore, the radius of the iron sphere is
$$
R=\frac{\| 3\left[9.30 \times 10^{-26} \mathrm{~kg} h \mid\left[8.0 \times 10^{22} \mathrm{~J} / \mathrm{T} \|\left.\right|^{1 / 3}\right]^{3 / 3}\right.}{\| 4 \pi\left[14 \times 10^3 \mathrm{~kg} / \mathrm{m}^3 \|\left[2.1 \times 10^{-23} \mathrm{~J} / \mathrm{T} \|\right.\right.}\\=1.8 \times 10^5 \mathrm{~m} .
$$
(b) The volume of the sphere is
$V_s=\frac{4 \pi}{3} R^3=\frac{4 \pi}{3}\left\{1.82 \times 10^5 \mathrm{~m}^{h^3}=2.53 \times 10^{16} \mathrm{~m}^3\right.$ and the volume of the Earth is
$$
V_E=\frac{4 \pi}{3} R_E^3=\frac{4 \pi}{3}\left(6.37 \times 10^6 \mathrm{~m}\right)^3\\=1.08 \times 10^{21} \mathrm{~m}^3,
$$
so the fraction of the Earth's volume that is occupied by the sphere is
$$
\frac{V_s}{V_E}=\frac{2.53 \times 10^{16} \mathrm{~m}^3}{1.08 \times 10^{21} \mathrm{~m}^3}\\=2.3 \times 10^{-5}
$$
