= Solution
Interpret the box as a hard-wall cube. Its one-particle levels are $\epsilon_{n_xn_yn_z}=\pi^2\hbar^2(n_x^2+n_y^2+n_z^2)/(2mL^2)$ for positive integers, giving $E_0=3\pi^2\hbar^2/(2mL^2)$. Counting lattice points in a positive octant and differentiating the large-radius count gives the supplied continuum <density of states>. Expressing its prefactor through $E_0$ yields
$$
g(\epsilon)=\mathcal A E_0^{-3/2}\sqrt\epsilon,\qquad
\boxed{\mathcal A=\frac{3\sqrt3\pi}{4}.}
$$
Here $\mathcal A$ denotes the dimensionless constant requested in this question, to distinguish it from the unrelated hyperfine constant of question 1. Periodic <boundary conditions> would have zero ground <energy> and would not permit this parametrization by a nonzero $E_0$.
Separate the ground-state occupation from excited particles. The <finite-box ground-energy convention in Bose condensation> makes the <chemical potential> approach the ground <energy> from below as that occupation becomes macroscopic. In the thermodynamic continuum approximation one shifts the <energy> zero to the ground state and sets the limiting <chemical potential> to zero in the excited-state integral. The difference between the exact shifted density and its $\sqrt\epsilon$ bulk form gives finite-size corrections; at the transition $k_BT_c/E_0\sim N^{2/3}\gg1$. Thus
$$
N'_{\max}(T)=\int_0^\infty\frac{\mathcal A E_0^{-3/2}\sqrt\epsilon}{e^{\epsilon/(k_BT)}-1}\,d\epsilon
=\mathcal A\Gamma(3/2)\zeta(3/2)\left(\frac{k_BT}{E_0}\right)^{3/2}.
$$
The identity follows by expanding $(e^x-1)^{-1}=\sum_{r\geq1}e^{-rx}$ and integrating each term. The excited states have a finite saturation capacity. Setting that capacity equal to $N$ determines condensation:
$$
\boxed{k_BT_c=\mathcal B E_0N^{2/3},\qquad
\mathcal B=[\mathcal A\Gamma(3/2)\zeta(3/2)]^{-2/3}
=\frac{4}{3\pi\zeta(3/2)^{2/3}}\simeq0.22375.}
$$
Equivalently $k_BT_c=(2\pi\hbar^2/m)[n/\zeta(3/2)]^{2/3}$, with $n=N/L^3$.
At fixed $N$, below $T_c$ the <chemical potential> is pinned to its limiting value and
$$
\boxed{N'=N(T/T_c)^{3/2},\qquad N_0=N[1-(T/T_c)^{3/2}].}
$$
Above $T_c$, $N'\simeq N$ and there is no macroscopic $N_0$; the <chemical potential> adjusts rather than allowing the saturation expression to exceed the available particles. At fixed $T$, put $N_c=N'_{\max}(T)$. For $N<N_c$, $N'\simeq N$ and $N_0$ is not macroscopic; for $N>N_c$, \b[$N'=N_c$ saturates while $N_0=N-N_c$ grows linearly]. This is <particle saturation in a homogeneous Bose gas>; the sharp threshold is a thermodynamic-limit approximation to finite-box occupations.
\Image[/past-exam-of-the-atomic-and-optical-physics-course-of-the-university-of-cambridge/2015/paper-1-bose-gas.png]
{title=Ideal-box particle saturation and the heat-capacity crossover from phonons to particle-like excitations}
The interaction scale compares differently with confinement and transition energies:
$$
\frac{E_I}{E_0}\sim\frac{Na}{L}\gg1,\qquad
\frac{E_I}{k_BT_c}\sim N^{1/3}\frac aL\ll1.
$$
Thus the stated range makes interactions dominate the ground-state confinement <energy> while remaining small on the thermal scale controlling $T_c$. This is an <interaction-dominated ground state with nearly ideal condensation temperature>. It also gives the dilute-gas parameter $na^3=N(a/L)^3\ll1$. A contact model has $g=4\pi\hbar^2a/m$ and $\mu\simeq gn$; the question's $E_I$ suppresses order-one factors, and the uniform ground <energy> per particle is $gn/2$, rather than exactly $gn$. Weak interactions can shift $T_c$ slightly, but their dimensionless strength $an^{1/3}\ll1$ makes that a small relative effect. In a hard-wall box, boundary healing layers do not overturn bulk interaction dominance when their thickness is small compared with $L$.
At $k_BT\ll\mu$, thermally occupied <Bogoliubov modes> have $E_k\simeq\hbar c_s k$, with $c_s=\sqrt{\mu/m}$. There is one acoustic branch for this spinless gas. Its thermal <energy> determines the <phonon heat capacity of a dilute Bose gas>:
$$
U_{\rm ph}=\frac{V}{2\pi^2}\int_0^\infty\frac{\hbar c_s k^3}{e^{\hbar c_sk/(k_BT)}-1}\,dk
=\frac{\pi^2V}{30}\frac{(k_BT)^4}{\hbar^3c_s^3},
$$
$$
\boxed{C_{V,\rm ph}=\frac{2\pi^2}{15}Vk_B\left(\frac{k_BT}{\hbar c_s}\right)^3\propto T^3.}
$$
The zero-temperature interaction <energy> does not itself contribute to this leading <temperature> derivative at fixed $N,V$. This bulk power law also requires $k_BT\gg\hbar c_s/L$ so that the <phonon> spectrum may be treated as continuous; a strictly finite box eventually has discrete-level activated behaviour.
For the noninteracting condensed <Bose gas>, the <ideal Bose-gas heat capacity below condensation> follows by integrating $\epsilon g(\epsilon)f(\epsilon)$:
$$
U=\frac32Nk_BT\frac{\zeta(5/2)}{\zeta(3/2)}(T/T_c)^{3/2},\qquad
\boxed{C_{V,\rm ideal}=\frac{15}{4}Nk_B\frac{\zeta(5/2)}{\zeta(3/2)}(T/T_c)^{3/2}\propto T^{3/2}.}
$$
Its limiting value at $T_c^-$ is about $1.926Nk_B$. At very low <temperature> the interacting <phonon> curve is suppressed relative to the ideal curve, has slope three on logarithmic axes, and crosses over towards particle-like $T^{3/2}$ behaviour around $k_BT\sim\mu$. Since $\mu\ll k_BT_c$, most of the interval below $T_c$ is approximately ideal at leading weak-coupling order. The figure marks a qualitative extrapolation near the transition: a zero-temperature <Bogoliubov spectrum> does not determine precise interacting critical <heat capacity> or its small transition shift. The illustrative fixed-$\mu$ interpolation is not claimed as a controlled critical-region calculation.
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