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The state marked X in F is the $1/2^{-}$ analogue at $11.20\mathrm{MeV}$ excitation. Its absolute nuclear energy relative to O is $2.25+11.20=13.45\mathrm{MeV}$. HenceIt is a same-spin, same-parity allowed beta decay between isobaric analog states. The large analogue Fermi overlap makes it a superallowed analogue transition, in the broad analogue-state terminology. Since $J=1/2$ rather than zero, this classification does not exclude a Gamow-Teller contribution.

For a transparent numerical estimate, take the Fermi part to dominate both this branch and the reference F-ground-state decay, neglect Coulomb and electron-mass corrections in their phase-space ratio, and use the Sargent rule. For $T_z=(N-Z)/2$, Ne has $T=T_z^{\mathrm{magnitude}}=3/2$ and goes from $T_z=-3/2$ to $-1/2$ in its analogue multiplet. The isospin ladder strength in analogue beta decay is therefore three. The F→O ground-state mirror transition in $T=1/2$ has Fermi strength one. Thus the analogue beta-decay branching estimate givesThis percent-level result is an estimate under the stated assumptions, not a unique consequence of the level diagram. More generally, if $f$ denotes an allowed phase-space integral and $S=g_V^2|M_F{|}^2+g_A^2|M_{GT}{|}^2$, thenThe diagram fixes the Fermi strengths but not both axial strengths. Including them and proper Positron phase-space functions changes the numerical result. Assuming equal total allowed strengths instead of the Fermi-only ratio three would give approximately $0.64\%$; that alternative makes a different nuclear-structure assumption. The rate enhancement from removing the forbiddenness explains why the X branch can be appreciable despite its much smaller endpoint than the Ne ground-state branch.