= Solution
Take one conduction <electron> per atom and a three-dimensional <free electron gas>, with <spin degeneracy> two and <electron> <mass> $m_e$. State counting inside the occupied <Fermi sphere> gives
$$
n=\frac2{(2\pi)^3}\frac{4\pi k_F^3}{3}=\frac{k_F^3}{3\pi^2}.
$$
The highest occupied <energy> at zero <temperature> is $E_F=\hbar^2k_F^2/(2m_e)$, so
$$
\boxed{T_F=\frac{\hbar^2}{2m_ek_B}(3\pi^2n)^{2/3}}.
$$
This relates the atomic <number density> to the <Fermi temperature> only because monovalency and the free-electron dispersion have been assumed. A real <energy band> can require a different <density of states> and <band effective mass>.
For the isotropic <Debye model>, count three acoustic <polarizations> up to one spherical <wavevector> cutoff $q_D$. The number of vibrational modes per <volume> is
$$
3\frac1{(2\pi)^3}\frac{4\pi q_D^3}{3}=3n,
\qquad q_D=(6\pi^2n)^{1/3}.
$$
With $\omega=cq$ and $\hbar\omega_D=k_B\theta_D$,
$$
\boxed{\theta_D=\frac{\hbar c}{k_B}(6\pi^2n)^{1/3}}.
$$
The effective $c$ is the <Debye sound speed>: matching the low-frequency mode density of one longitudinal and two transverse branches requires $3/c^3=1/c_L^3+2/c_T^3$. It is not generally the arithmetic average of their speeds. The cutoff preserves the mode count; it need not coincide with the actual nonspherical <Brillouin zone> boundary.
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