= Solution
The <weak charged current> creates a negative charged <lepton> in a left-chiral state. For a massless particle, that would have negative <helicity>; part (e) instead requires positive <helicity>. A nonzero <mass> supplies the necessary wrong-helicity component. This <helicity suppression> is extremely strong for the fast <Electron>, but much weaker for the slow <muon>.
For a positive-helicity charged lepton with spinors normalized to unit probability, the left-chiral weight is $(1-\beta_\ell)/2$. With the box-state normalization of part (c), this gives the useful consistent rate ratio
$$
R=\frac{\rho_\mu}{\rho_e}\frac{1-\beta_\mu}{1-\beta_e}.
$$
Indeed $1-\beta_\ell=2m_\ell^2/(m_\pi^2+m_\ell^2)$ cancels the extra energy-dependent factor in that density ratio, leaving the standard <pion electron-to-muon decay ratio> in inverted form:
$$
\boxed{R=\frac{m_\mu^2}{m_e^2}
\left[\frac{1-m_\mu^2/m_\pi^2}{1-m_e^2/m_\pi^2}\right]^2.}
$$
Using the rounded masses in this paper gives $R\simeq8185$, close to the stated order of $8000$. Precision predictions require more accurate <masses> and electromagnetic radiative corrections. The essential mechanism is the charged-lepton-mass dependence of the <decay amplitude>; simply counting available final <momenta> misses it.
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