What is the Fermi energy of Liquid He3 at absolute zero?

[atomicpedals]

Homework Statement

The atom $He^3$ has spin 1/2 and is a fermion. The density of liquid $He^3$ is $0.081g/cm^3$ near absolute zero. Calculate the Fermi energy $\epsilon_F$ and the Fermi temperature $T_F$

Homework Equations

$\epsilon_F = \frac{\hbar^2}{2m}(\frac{3 \pi^2 N}{V})^{2/3}$

$T_F = \frac{\epsilon_F}{k}$

The Attempt at a Solution

In the problem I'm given the density is $0.081g/cm^3$, which is my N/V. Assuming that m is the mass of $He^3$, then $m = 5.008*10^{-24}g$. I should then be able to do a straight forward plug-n-chug; however my units don't work out as I get:

$\epsilon_F = \frac{\hbar^2}{1.0016*10^{-23}g}(3 \pi^2 0.081g cm^{-3})^{2/3}=1.99*10^{-50} kg^{5/3} m^2 s^{-2}$

Clearly, this is not a unit of energy. What am I doing wrong?

 

[atomicpedals]

I need to use number density, rather than straight density. Which should take care of my problem.