= Solution
In the one-band electron <Hall effect> model, $R_H=-1/(ne)$, with $e>0$. The measured <Hall coefficient> therefore gives
$$
n=\frac1{e|R_H|}=2.4966\times10^{28}\,\mathrm{m^{-3}}.
$$
A monovalent <body-centered cubic lattice> has two conduction electrons per conventional <unit cell>. Thus $na^3=2$, and
$$
\boxed{a=\left(\frac2n\right)^{1/3}=4.3108\times10^{-10}\,\mathrm m.}
$$
Counting two spin states within the <Fermi sphere> gives $k_F=(3\pi^2n)^{1/3}=9.0418\times10^9\,\mathrm{m^{-1}}$. Using the specified <effective mass> in the dispersion,
$$
\boxed{E_F=\frac{\hbar^2k_F^2}{2m^*}=2.3960\,\mathrm{eV},\qquad
T_F=E_F/k_B=2.7805\times10^4\,\mathrm K.}
$$
There is one mole of conduction electrons per mole of sodium. Its molar <density of states> at $E_F$ is $3N_A/(2E_F)$, giving
$$
\boxed{\gamma_{\mathrm{mol}}=\frac{\pi^2N_Ak_B^2}{2E_F}
=1.4757\,\mathrm{mJ\,mol^{-1}\,K^{-2}}.}
$$
Using the volume <density of states> in <2/c> gives
$$
\boxed{\chi_p=1.0543\times10^{-5}\quad\text{(dimensionless SI volume susceptibility)}.}
$$
The <effective mass> changes the dispersion and <density of states>; the electron <Bohr magneton> used for the spin coupling retains the bare electron mass. Replacing it by $e\hbar/(2m^*)$ would be an additional, unjustified magnetic-moment assumption. A Hall factor or a multiband metal would also invalidate the simple density inference; the stated calculation uses the requested one-band approximation.
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