Answer
0.09%
Work Step by Step
Let's apply the ideal gas law $PV=nRT$ to the gas to find the volume of the gas.
$PV=nRT=>V_{gas}=\frac{nRT}{P}-(1)$
We can write the desired percentage as follows,
Percentage = $\frac{V_{atoms}}{V_{gas}}\times100\%$
Percentage = $\frac{(\frac{4}{3}\pi r^{3})nN_{A}}{V_{gas}}\times100\%-(2)$
(1)=>(2),
Percentage = $\frac{(\frac{4}{3}\pi r^{3})nN_{A}}{\frac{nRT}{P}}\times100\%=\frac{(\frac{4}{3}\pi r^{3})N_{A}P}{RT}\times100\%$
Let's plug known values into this equation.
Percentage = $\frac{x\frac{4}{3}\pi(2\times10^{-10}m)^{3}(6.022\times10^{23}mol^{-1})(1.01\times10^{5}Pa)}{(8.31\space J/mol\space K)(273\space K)}\times100\%=0.09\%$