= Solution
In the spin-degenerate <free electron model>, a monovalent metal has conduction-electron <number density> $n$. The density of wave-vector states is $1/(2\pi)^3$ per unit real-space <volume>, with two <spin> states at each <wave vector>. Filling a sphere at zero <temperature> gives
$$
n=\frac{2}{(2\pi)^3}\frac{4\pi k_F^3}{3}=\frac{k_F^3}{3\pi^2},\qquad
\boxed{T_F=\frac{E_F}{k_B}=\frac{\hbar^2}{2m_ek_B}(3\pi^2n)^{2/3}.}
$$
For the <Debye model>, there are three acoustic polarizations and $3n$ modes per unit <volume> in total. Replacing their dispersions by $\omega=ck$ with the effective <Debye sound speed>, mode counting gives
$$
3\frac{1}{(2\pi)^3}\frac{4\pi k_D^3}{3}=3n,
\qquad k_D=(6\pi^2n)^{1/3},
\qquad \boxed{\theta_D=\frac{\hbar c}{k_B}(6\pi^2n)^{1/3}.}
$$
The factor two in the <electron> count is <spin>, whereas the factor three in the <phonon> count is polarization; the two counts are not interchangeable. If branch speeds differ, the Debye effective average obeys $c^{-3}=(c_l^{-3}+2c_t^{-3})/3$ in the isotropic approximation.
Dividing gives $T_F/\theta_D=\lambda(3\pi^2n/2)^{1/3}$, where $\lambda=\hbar/(2m_ec)$. A monatomic FCC conventional cell contains four atoms, so $n=4/a^3$. Therefore
$$
\boxed{\frac{T_F}{\theta_D}=(6\pi^2)^{1/3}\frac\lambda a.}
$$
Here and below $c$ is the \b[speed of sound], not the speed of light. The Debye cutoff counts acoustic modes in a spherical approximation to the <Brillouin zone>; it is not an electron-state cutoff.
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