Calculate the radius of an Ag atom
If you had included the units in your calculation, you would have noticed why your equation is not correct.

Molar mass $M$ is defined as
$$M=\frac mn\tag1$$
where $m$ is mass and $n$ is amount of substance.
Since the Avogadro constant $N_\mathrm A$ is
$$N_\mathrm A=\frac Nn\tag2$$
where $N$ is the number of particles, the mass $m$ of one atom $(N=1)$ is
$$m=\frac M{N_\mathrm A}\tag3$$

Density $\rho$ is defined as 
$$\rho=\frac mV\tag4$$
where $V$ is volume.
Thus, the volume of a sample is
$$V=\frac m\rho\tag5$$
Using Equation $\text{(3)}$, the volume $V$ can be calculated for a single atom:
$$V=\frac M{N_\mathrm A\cdot\rho}\tag6$$

Assuming that a fraction of $88\,\%$ of the volume $V$ is filled with a hard sphere, the volume $V_\text{sphere}$ of the sphere is
$$\begin{align}
V_\text{sphere}&amp;=0.88\times V\tag7\\[6pt]
&amp;=0.88\times\frac M{N_\mathrm A\cdot\rho}\tag8
\end{align}$$

Since the volume of a sphere is
$$V_\text{sphere}=\frac43\pi r^3\tag9$$
where $r$ is the radius of the sphere, the radius $r$ is
$$\begin{align}
r&amp;=\sqrt[3]{\frac{3V_\text{sphere}}{4\pi}}\tag{10}\\[6pt]
&amp;=\sqrt[3]{\frac{3\times0.88\times M}{4\pi\cdot N_\mathrm A\cdot\rho}}\tag{11}\\[6pt]
&amp;=\sqrt[3]{\frac{3\times0.88\times 107.86820\ \mathrm{g\ mol^{-1}}}{4\pi\times
6.02214076\times10^{23}\ \mathrm{mol^{-1}} \times 10.5\ \mathrm{g\ cm^{-3}}}}\\[6pt]
&amp;=1.53\times10^{-8}\ \mathrm{cm}\\[6pt]
&amp;=1.53\times10^{-10}\ \mathrm m\\[6pt]
&amp;=153\times10^{-12}\ \mathrm m\\
\end{align}$$