= Solution
There is a $t$-channel <Z boson> contribution and a crossed, $u$-channel <W boson> contribution. The latter pairs the incoming <electron neutrino> with the outgoing <electron>, and the incoming <electron> with the outgoing <electron neutrino>. Both contribute coherently to the same elastic final state.
\Image[/past-exam-of-the-nuclear-and-particle-physics-course-of-the-university-of-oxford/2022-neutrino-diagrams.png]
{title=Neutral- and charged-current diagrams for the four elastic neutrino scattering channels}
For a concrete comparison, take unpolarized targets, the same <Mandelstam invariant> $s$ in all four reactions, and $m_e^2\ll s\ll m_W^2$. In a stationary-<electron> experiment this includes $E_\nu\gg m_e$ with $s\simeq2m_eE_\nu$. Define $x=\sin^2\theta_W$, using the <weak mixing angle>, and $\sigma_*=G_F^2s/\pi$. The <weak neutral current> has chiral couplings $g_L=x-1/2$, $g_R=x$; the <weak charged current> shifts $g_L$ by one. Consequently
$$
\boxed{\sigma_{\mathrm{i}}=\sigma_*\left[(x+\tfrac12)^2+\frac{x^2}{3}\right].}
$$
The factor $1/3$ comes from integrating the opposite-chirality angular weight: in the massless limit $\int_0^1(1-y)^2dy=1/3$. Thus the relative rate depends on the chiral couplings and their <interference>, rather than simply on the number of diagrams.
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