= Solution
In the simple box-state counting used here, the final <three-momentum> lies in a spherical shell: $dn=Cp^2dp$, where the common constant $C$ contains the normalization volume and angular factor. <Momentum conservation> fixes the other final particle's <momentum>. Hence the <density of states> is
$$
\rho_\ell(E_0)=Cp_\ell^2\frac{dp_\ell}{dE_0}.
$$
Combining parts (a) and (b) gives
$$
\boxed{\frac{\rho_\mu}{\rho_e}
=\left[\frac{m_\pi^2-m_\mu^2}{m_\pi^2-m_e^2}\right]^2
\frac{m_\pi^2+m_\mu^2}{m_\pi^2+m_e^2}.}
$$
With the paper's rounded <masses>,
$$
R_{\mathrm{naive}}=\left(\frac{19600-11236}{19600-0.25}\right)^2
\frac{19600+11236}{19600+0.25}
=0.28650.
$$
By <Fermi's golden rule>, equal squared matrix elements make the rate ratio equal to this density ratio. Thus \b[the naive constant-matrix-element estimate is $R\simeq0.287$], which favours the electronic decay by about $3.49$ to one.
This is the particular noncovariant state-counting convention in the question. A calculation with the <Lorentz-invariant phase-space measure> places factors $1/(2E)$ in the measure and correspondingly normalizes its <decay amplitude> differently. Only the complete product of phase space and squared matrix element is convention-independent; mixing the two conventions would give a wrong rate.
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