= Solution
Take <metric> $g_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$ and write $M=M_W$. The outgoing electron and <antineutrino> momenta are $k$ and $k'$. Up to an irrelevant overall phase, the <amplitude> is
$$
\mathcal M=\frac{g}{2\sqrt2}\epsilon_\mu\bar u(k)\gamma^\mu(1-\gamma_5)v(k').
$$
A <fermion spin sum> gives
$$
\sum_{\rm final}|\mathcal M|^2=\frac{g^2}{8}\epsilon_\mu\epsilon_\nu^*T^{\mu\nu},\qquad T^{\mu\nu}=\operatorname{Tr}\bigl[\not k\gamma^\mu(1-\gamma_5)\not k'\gamma^\nu(1-\gamma_5)\bigr].
$$
Anticommuting $\gamma_5$ through the massless <momentum> and <gamma matrix>, and using $(1-\gamma_5)^2=2(1-\gamma_5)$, reduces the <trace> to twice the four-gamma <trace> minus twice its $\gamma_5$ counterpart. In particular its symmetric part is
$$
T_S^{\mu\nu}=8\bigl(k^\mu k'^\nu+k^\nu k'^\mu-g^{\mu\nu}k\cdot k'\bigr).
$$
The epsilon-tensor part is antisymmetric and contributes neither to the unpolarized polarization sum nor to the real linear polarization used below. There is no extra factor for selecting the <neutrino> <helicity>: the chiral vertex has already done so.
The external massless equations imply $p_\mu J^\mu=0$ for $p=k+k'$, so $p_\mu T^{\mu\nu}=0$. With $k\cdot k'=M^2/2$, averaging over the three initial <spin> states gives
$$
\overline{|\mathcal M|^2}=\frac{g^2}{24}\left(-g_{\mu\nu}+\frac{p_\mu p_\nu}{M^2}\right)T^{\mu\nu}=\boxed{\frac{g^2M^2}{3}}.
$$
Indeed $-g_{\mu\nu}T_S^{\mu\nu}=16k\cdot k'=8M^2$. This is a <massive-vector spin average>, not an average over the final <spins>.
In the rest frame, eliminating the <momentum> delta function and integrating the remaining radial delta function $\delta(M-2|\mathbf k|)$ reduces the <two-body Lorentz-invariant phase space> to $d\Phi_2=d\Omega/(32\pi^2)$. Therefore
$$
\frac{d\Gamma}{d\Omega}=\frac{|\mathcal M|^2}{64\pi^2M},\qquad\boxed{\Gamma_{\rm unpol}=\frac{g^2M}{48\pi}}.
$$
This derives the <massless leptonic W decay width> without invoking a precomputed decay formula.
For the specified polarization there is no initial <spin average>. Put $\mathbf k=(M/2)\widehat{\mathbf n}$, $\mathbf k'=-(M/2)\widehat{\mathbf n}$ and $\widehat n_z=\cos\theta$. The contraction selects
$$
T_S^{33}=8\left(-\frac{M^2}{2}\cos^2\theta+\frac{M^2}{2}\right)=4M^2\sin^2\theta.
$$
It follows that
$$
\boxed{\sum_{\rm final}|\mathcal M_z|^2=\frac{g^2M^2}{2}\sin^2\theta,\quad\frac{d\Gamma_z}{d\Omega}=\frac{g^2M}{128\pi^2}\sin^2\theta,\quad\Gamma_z=\frac{g^2M}{48\pi}.}
$$
Here $\int\sin^2\theta\,d\Omega=8\pi/3$. The total rate agrees with the spin-averaged result, but its normalized polar distribution is $(1/\Gamma_z)d\Gamma_z/d\cos\theta=3(1-\cos^2\theta)/4$ instead of the isotropic unpolarized distribution. This is <polarized massless leptonic W decay>.
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