= Solution
The <tree-level Feynman diagram> contains one <weak charged current> vertex. Fermion arrows point along fermion-number flow, so the outgoing <antiquark> arrow points toward the vertex.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-305-w-decay.png]
{title=Tree-level W-plus decay into an outgoing quark and antiquark with fermion-flow arrows}
For one fixed matching color, the <scattering amplitude> is
$$
\mathcal M=\frac{g}{2\sqrt2}V_{qq'}\epsilon_\mu(p)\bar u(k)\gamma^\mu(1-\gamma^5)v(k').
$$
An overall phase from the vertex has no effect on the width. The <fermion spin sums> give $\sum u\bar u=\not k$, $\sum v\bar v=\not k'$ in the massless approximation. For the three initial polarizations, use a <spin average> of $1/3$. The <gamma matrix trace identities> give
$$
T^{\mu\nu}=\operatorname{Tr}\left[\not k\gamma^\mu(1-\gamma^5)\not k'\gamma^\nu(1-\gamma^5)\right]
=8\left(k^\mu k'^\nu+k^\nu k'^\mu-g^{\mu\nu}k\cdot k'+i\epsilon^{\mu\nu\rho\sigma}k_\rho k'_\sigma\right).
$$
The sign of the last term follows the PDF's $\gamma^5$ convention and does not affect this decay: the polarization sum is symmetric. Since $p=k+k'$ and the daughter masses vanish, $p_\mu T^{\mu\nu}=0$ and $k\cdot k'=M_W^2/2$. Therefore
$$
\overline{|\mathcal M|^2}_{\text{one color}}
=\frac{g^2|V_{qq'}|^2}{24}\left(-g_{\mu\nu}+\frac{p_\mu p_\nu}{M_W^2}\right)T^{\mu\nu}
=\frac{g^2M_W^2}{3}|V_{qq'}|^2.
$$
In the rest frame, the two-body <Lorentz-invariant phase space> integrates to $1/(8\pi)$: after the spatial delta function sets $\mathbf k'=-\mathbf k$, its radial delta function fixes $|\mathbf k|=M_W/2$. Explicitly,
$$
\int d\Phi_2=\frac{1}{16\pi^2}\int d\Omega\int_0^\infty dk\,\delta(M_W-2k)=\frac{1}{8\pi}.
$$
The decay formula consequently gives
$$
\Gamma_{\text{one color}}=\frac{\overline{|\mathcal M|^2}}{16\pi M_W}
=\frac{g^2M_W}{48\pi}|V_{qq'}|^2
=\boxed{\frac{G_FM_W^3}{6\pi\sqrt2}|V_{qq'}|^2.}
$$
\b[This is the printed expression, interpreted for one color.] For a physical quark-flavor channel there are three orthogonal final color states. Their probabilities add; no initial color average is present for a colorless <W boson>. Thus
$$
\boxed{\Gamma_{W^+\to q\bar q'}=N_c\frac{G_FM_W^3}{6\pi\sqrt2}|V_{qq'}|^2,\qquad N_c=3.}
$$
The PDF's formula omits this color multiplicity if read as the ordinary inclusive flavor width. It is also the familiar normalization for a colorless lepton channel.
The six physically accessible <quark> combinations are \b[$u\bar d,u\bar s,u\bar b,c\bar d,c\bar s,c\bar b$]. A real on-shell <W boson> cannot produce a <top quark>; the massless approximation is applied to the accessible daughters and does not open a physically forbidden top channel. <CKM matrix> unitarity gives $\sum_{q'=d,s,b}|V_{uq'}|^2=\sum_{q'=d,s,b}|V_{cq'}|^2=1$. Therefore the physical color-summed hadronic width at this order is
$$
\boxed{\Gamma_{\rm had}=\frac{G_FM_W^3}{\pi\sqrt2}.}
$$
If the printed one-color convention is retained for every channel, its sum is instead $G_FM_W^3/(3\pi\sqrt2)$. In the artificial theory where all three up-type flavors, including the top, are kinematically massless, there would be nine channels and the physical sum would be $9G_FM_W^3/(6\pi\sqrt2)$; that is not the on-shell Standard Model channel list.
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