= Solution
For a <nucleon> at rest and beam <energy> well above its <mass> scales, $s\simeq2m_NE$. If the leading <Parton distribution functions>[parton distributions] are approximately independent of beam <energy>, integration over $x,y$ therefore makes both total <cross sections> approximately proportional to $E$. The plotted plateaus are evidence for scattering on approximately pointlike constituents in this range, with the departures expected from <QCD> evolution and threshold effects.
Define the momentum-weighted contents $Q=\int_0^1xq(x)dx$ and $\bar Q=\int_0^1x\bar q(x)dx$. Since $\int_0^1(1-y)^2dy=1/3$, the two <cross sections> obey
$$
\frac{\sigma_\nu}{E}\propto Q+\frac{\bar Q}{3},\qquad
\frac{\sigma_{\bar\nu}}{E}\propto\frac Q3+\bar Q.
$$
Read the original high-energy plateaus as approximately $0.67$ and $0.33$ in the plotted $10^{-38}\,\mathrm{cm^2/GeV}$ units. With $R=\sigma_{\bar\nu}/\sigma_\nu\simeq1/2$ and $r=\bar Q/Q$,
$$
R=\frac{1/3+r}{1+r/3},\qquad
\boxed{r=\frac{3R-1}{3-R}\simeq0.20.}
$$
Using the literal readings $0.33/0.67$ gives $r=0.1905$, so twenty percent is an appropriate graphical estimate, not a precision fit. A purely valence target would have $R=1/3$, while a fully symmetric sea would give $R=1$.
The <neutrino-antineutrino DIS sea momentum ratio> is a ratio of <momentum> fractions in the effective <Flavor eigenstate>[flavour] combinations, not an unweighted count of <antiquarks>. The cross-section integral contains $xq(x)$: it cannot determine $\int\bar q/\int q$ without additional assumptions about their shapes. A distribution growing at small $x$ can have a large number integral while its <momentum> integral remains finite.
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