= Solution
As $T\downarrow0$, the <Fermi-Dirac distribution> tends to a step function: $\bar n(p)=1$ for $\epsilon(p)<\mu_0$, and $0$ for $\epsilon(p)>\mu_0$. The value at equality is immaterial to the integrals. The <Fermi momentum> is defined by $\epsilon(p_F)=\mu_0$: the <Pauli exclusion principle> fills the lowest available states and forbids multiple occupation of a state with a fixed spin. Thus the sketch is a horizontal line at one up to $p_F$, followed by a jump to zero.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ii/paper-3-fermi-step.png]
{title=Zero-temperature Fermi occupation as a function of momentum; the value at the jump is immaterial}
$$
\boxed{n=\frac{4\pi g_s}{3h^3}p_F^3.}
$$
In the <ultrarelativistic>, $\epsilon(p)\simeq cp$, giving
$$
\frac EV\simeq\frac{4\pi g_s c}{h^3}\int_0^{p_F}p^3\,dp=\frac{\pi g_sc}{h^3}p_F^4=\frac34ncp_F.
$$
Eliminate $p_F$ to obtain
$$
\boxed{E\simeq\frac34\left(\frac3{4\pi g_s}\right)^{1/3}hc\,n^{4/3}V\sim hc\,n^{4/3}V.}
$$
The supplied expression for $\epsilon$ includes rest energy; in this limit its leading term is also the leading kinetic energy.
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