= Solution
Read the <spin> and <parity> from the actual diagram: N and Ne have $1/2^-$ ground states; O and F have $5/2^+$ ground states. The N→O beta-minus branch and Ne→F beta-plus branch therefore both have $|\Delta J|=2$ with a <parity> change: \b[both are first-forbidden unique]. The F→O beta-plus branch has $5/2^+\to5/2^+$: \b[it is allowed and can contain both Fermi and Gamow-Teller contributions].
The Ne ground-state branch is \b[$0.2\%=0.002$], as explicitly printed on the diagram, not $0.2$ as a probability. The prose leaves off its percent sign. Interpret the supplied “lifetimes” as mean lifetimes $\tau$, so each partial decay rate is $\lambda_j=B_j/\tau$, with no extra $\ln2$. Then
$$
\boxed{\lambda_{N\to O}=\frac{0.016}{6.1}=2.623\times10^{-3}\ \mathrm{s^{-1}},}
$$
$$
\boxed{\lambda_{F\to O}=\frac1{95}=1.053\times10^{-2}\ \mathrm{s^{-1}},\qquad
\lambda_{Ne\to F}=\frac{0.002}{0.15}=1.333\times10^{-2}\ \mathrm{s^{-1}}.}
$$
A <decay width> in <energy> units is $\Gamma_j=\hbar\lambda_j$, rather than a rate. Thus the corresponding partial <energy> widths, in the same order N, F, Ne, are \b[$1.73\times10^{-24}$, $6.93\times10^{-24}$ and $8.78\times10^{-24}\ \mathrm{MeV}$]. If a different convention used the supplied numbers as half-lives, all these rates and widths would acquire $\ln2$; their ratios would be unchanged. The literal mean-lifetime convention is used throughout.
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