= Solution
Set the ground-state <energy> to zero. Bose-Einstein occupation of an excited level is $[e^{(\epsilon-\mu)/(k_BT)}-1]^{-1}$, with $\mu\le0$. As particles are added, $\mu$ approaches zero and the excited-state population saturates. Separating the ground state from the density-of-states integral therefore gives
$$
\boxed{N_c(T)=\int_0^\infty\frac{g(\epsilon)\,d\epsilon}{e^{\epsilon/(k_BT)}-1},
\qquad\frac{N_0}{N}=1-\frac1\eta.}
$$
The excited-state integral must actually converge: $g\to0$ alone is not a sufficient condition in complete generality. A <power> law $g\propto\epsilon^s$ with $s>0$ suffices at low <energy>, as in the stated three-dimensional box.
Let $\mathcal C=Vm^{3/2}/(\sqrt2\pi^2\hbar^3)$ denote the density-of-states prefactor, not a <heat capacity>. Substitution $x=\epsilon/(k_BT)$ gives
$$
N_c=\mathcal C(k_BT)^{3/2}\Gamma(3/2)\zeta(3/2)
=\boxed{\zeta(3/2)\frac V{\lambda^3}}.
$$
The condensate has zero <energy> on this reference, so a further factor of $\epsilon$ in the integral gives
$$
E=\mathcal C(k_BT)^{5/2}\Gamma(5/2)\zeta(5/2),
\qquad
\boxed{\frac EN=\frac32\frac{N_c}{N}
\frac{\zeta(5/2)}{\zeta(3/2)}k_BT.}
$$
Relative to the classical value, the ratio is $\zeta(5/2)/[\eta\zeta(3/2)]\simeq0.5134/\eta<0.5134$. Macroscopic occupation of the zero-energy state and the Bose distribution both lower the average <energy>.
At fixed volume the condensed gas has $E\propto T^{5/2}$. Unbiased random loss keeps average $E/N$ constant, hence
$$
\boxed{T\propto N^{2/5}.}
$$
Removing a condensate atom removes essentially no <energy> and by itself leaves the excited cloud's <temperature> unchanged. Removing an excited atom removes its <energy>; during re-equilibration, some zero-energy condensate atoms join the excited population and share the remaining <energy>, lowering <temperature>. Averaging over unbiased loss gives the stated constant <energy> per particle, even though each individual type of loss need not preserve it separately.
Using $m\simeq87u$ and the supplied volume,
$$
\lambda_i\simeq0.837\,\mathrm{\mu m},\quad
N_c(T_i)\simeq1.78\times10^5,\quad
\boxed{\eta_i\simeq1.68.}
$$
There is only one occupied internal <spin> state, so no factor $2F+1$ belongs in the <density of states>. During loss, $\eta\propto N/T^{3/2}\propto N^{2/5}$, hence $\eta/\eta_i=T/T_i$. Condensation disappears at $\eta_f=1$:
$$
\boxed{T_f=\frac{T_i}{\eta_i}\simeq29.7\,\mathrm{nK}.}
$$
As a consistency check, $N_f=N_i\eta_i^{-5/2}\simeq8.15\times10^4$ agrees with $N_c(T_f)$.
Both this process and <evaporative cooling of atoms> require particle loss and re-equilibration. Ordinary evaporation preferentially removes energetic atoms, reducing $E/N$, and can cool an uncondensed classical gas. Here loss is <energy> independent, $E/N$ stays constant, and cooling relies on the condensed gas's unusual <equation of state>; random loss of a classical <ideal gas> would leave its <temperature> unchanged. Exact noninteraction would prevent collisional rethermalization: the imposed equilibrium assumption can be understood as weak thermalizing interactions whose equation-of-state corrections are neglected.
\Image[/past-exam-of-the-atomic-and-optical-physics-course-of-the-university-of-cambridge/2014/paper-1-aop-random-loss.png]
{title=Temperature and condensate fraction during unbiased loss, ending at the condensation threshold}
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