Answer
$a.\quad 3.0\times 10^{23}\ \mathrm{molecules}/\mathrm{m}^{3}$
$b.\quad$ greater
$c.\quad $ With $\mathrm{T}_{\mathrm{E}}\approx 290\mathrm{K},\quad (\displaystyle \frac{N}{V})_{\mathrm{E}}=2.52\times 10^{25}\ \mathrm{molecules}/\mathrm{m}^{3}$
The estimate is about 84 times that of Mars.
(These are rough, approximate values.)
Work Step by Step
Use $ \quad PV=NkT \quad (17-2)$ to express $\displaystyle \frac{N}{V}=\frac{P}{kT}$
(number of molecules per volume),
$k =1.38\times 10^{-23}\ \mathrm{J}/\mathrm{K}$ is Boltzmann's constant.
---
The temperature needs to be in kelvins.
$T_{\mathrm{c}}=\displaystyle \frac{5}{9}(T_{\mathrm{f}}-32)=\frac{5}{9}(-64-32)=-53.3^{\mathrm{o}}\mathrm{C}$
$T_{\mathrm{k}}=(273.15-53.3\mathrm{K})=219.8\mathrm{K}$
$a.$
$\displaystyle \frac{N}{V}=\frac{P}{kT}=\frac{0.92\times 10^{3}\ \mathrm{P}\mathrm{a}}{(1.38\times 10^{-23}\ \mathrm{J}/\mathrm{K})(219.8\ \mathrm{K})}=3.0\times 10^{23}\ \mathrm{molecules}/\mathrm{m}^{3}$
$b.$
Comparing $ \displaystyle \frac{N}{V}=\frac{P}{kT}$ on Earth and on Mars,
After looking up the mean temperatures, Earth's is about 290K ($14^{\mathrm{o}}\mathrm{C})$.
So in terms of kelvins, $\mathrm{T}_{\mathrm{E}}\approx 1.3\mathrm{T}_{\mathrm{M}}.$
The atmospheric pressure on Earth, however, is far greater than that on Mars.
So, the ratio $\displaystyle \frac{N}{V}$ is far greater on Earth than it is on Mars.
$c.$
With mean temperature on Earth $\mathrm{T}_{\mathrm{E}}\approx 290\mathrm{K},$ and pressure of $1.01\times 10^{5}\mathrm{P}\mathrm{a},$
$\displaystyle \frac{N}{V}=\frac{P}{kT}=\frac{1.01\times 10^{5}\mathrm{P}\mathrm{a}}{(1.38\times 10^{-23}\mathrm{J}/\mathrm{K})(290\mathrm{K})}=2.52\times 10^{25}\ \mathrm{molecules}/\mathrm{m}^{3}$
which is about $84$ times that of Mars.
