The question I have been given is:
Silver atoms in a metallic lattice only fill up $88\,\%$ of the space ($12\,\%$ is empty). The density of silver is $10.5\ \mathrm{g\cdot cm^{-3}}$. Assuming that silver atoms are hard spheres ($V=\tfrac43\cdot\pi\cdot r^3$, when $r$ is atomic radius), what is the radius of a silver atom? Give the answer in units of $10^{-12}$ meters.
The atomic mass of $\ce{Ag}$ is 107.8682.
My solution:
$$V=0.88\times V$$
$$V=\frac{0.88\times10.5\times6.022\times10^{23}}{107.8682}=5.158\times10^{22}\ \mathrm{cm^3}$$
$$V=\frac43\cdot\pi\cdot r^3 \Rightarrow r=\left(\frac34\cdot\frac V\pi\right)^{1/3}$$
Then I switched to the $10^{12}$ meters, the result was $4.953\times10^{17}$ and it is not correct. What am I doing wrong?