Proposed solution:
This question comes from Tanner's manual Introduction to the Physics of Electrons in Solids, at the chapter dedicated to the application of the Fermi gas model. The Fermi energy is $\mathsf{E}_F$. The model used here is the free electron model, and the susceptibility is $$ \mathcal{\chi} = \frac{\bf \mathcal{M}}{\bf B} $$
Assuming that the density of states at the Fermi level $D(\mathsf{E}_F)$ is approximated to: $$D(\mathsf{E}_F) = \frac{3N}{2} \mathsf{E}_F $$ where $N$ is the total number of electronic states. It is demonstrated that: $$ \mathcal{\chi}=\frac{3n\mu_B^2}{2\mathsf{E}_F}$$ where $n$ is the number of conduction electrons per unit volume given to be $n=2.5 \times 10^{28}\,m^{-3}$, $\mu_B$ is the Bohr magneton given to be $\mu_B = 9.3 \times 10^{-24}\,J\cdot T^{-1}$, the Fermi energy for $\ce{Na}$ is $\mathsf{E}_F = 5.2 \times 10^{-19}\,J$, and therefore the susceptibility is $\mathcal{\chi}=6.2\times 10^1\, MKS$, which can be converted in SI by multiplication of $4\pi \times 10^{-7}$, so the result is $\mathcal{\chi} = 7.8\times 10^{-5} \, SI$
Now my question:
This result is absolutely not consistent with the literature that states $$\mathcal{\chi} = 1.3\times 10^{-5} \, SI$$
What is the problem with Tanner or (most certainly) with my calculus ?
