#### Answer

a) $1.143\times10^{21}$ atoms, $76.4\%$ of the volume.
b) 4

#### Work Step by Step

a) Number of atoms: $cube's\ mass/atom's\ mass=0.1206/(1.055\times10^{-22})=1.143\times10^{21}$
Volume of one atom ($r=128\times10^{-12}\ m$): $V_a=\frac43\pi r^3=8.78\times10^{-30}\ m^3=8.78\times10^{-24}\ cm^3$
Volume of that number of atoms: $V_m=1.00\times10^{-2}\ cm^3$
Volume of the cube ($L=0.236\ cm$): $V_t=L^3=1.31\times10^{-2}\ cm^3$
Fraction of the cube which is occupied by atoms: $V_m/V_t\times100\%=76.4\%$
Spheres touching each other leave room "unutilized" due to their geometry.
b) Volume of the cube ($L=361.47\times10^{-10}\ cm$): $V=L^3=4.7230\times10^{-23}\ cm^3$
With the density of 8.960 g/cm³, the mass of the cube is $m_c=4.232\times10^{-22}\ g$
Since the mass of an atom is $1.055\times10^{-22}$, there are $4.011\approx 4$ atoms in the cube.