Answer
Please see the work below.
Work Step by Step
We know that
$n=\frac{\frac{4}{3}\pi(\frac{D_{cell}}{2})^3}{\frac{4}{3}\pi(\frac{D_{atom}}{2})^3}$
$n=(\frac{D_{cell}}{D_{atom}})^3$
We plug in the known values to obtain:
$n=(\frac{10\times 10^{-6}}{0.1\times 10^{-9}})^3$
$n=(10^5)^3=10^{15}$
Now the number of atoms in the body can be determined as
$n^{\prime}=(10^{15})(10^{15})\approx10^{30}atoms$
