= Solution
Use $f=X_{22}$ for the neon-22 <mass> fraction, temporarily keeping it distinct from the earlier metallicity. Let $n_b=\rho/m_u$ be the baryon density, ignoring small nuclear <mass> defects. Equal masses of <carbon> and <oxygen> give
$$
n_C=\frac{1-f}{24}n_b,\qquad n_O=\frac{1-f}{32}n_b,\qquad
n_{22}=\frac f{22}n_b.
$$
Carbon-12 and oxygen-16 each have equal <proton> and <neutron> numbers, while neon-22 has ten <protons> and twelve <neutrons>. The total <proton> and <neutron> numbers per baryon are consequently
$$
Y_e=6\frac{1-f}{24}+8\frac{1-f}{32}+10\frac f{22}
=\frac12-\frac f{22},
$$
$$
1-Y_e=\frac12+\frac f{22}.
$$
Hence
$$
\boxed{\frac{\langle Z\rangle}{\langle N\rangle}
=\frac{Y_e}{1-Y_e}=\frac{11-f}{11+f}.}
$$
The equal <carbon>/<oxygen> <mass> split cancels from this result: any mixture of these two symmetric nuclei has the same electron fraction. Taking the paper's metallicity symbol to mean $f$ gives its requested expression.
The literal pre-helium-burning <nitrogen> interpretation, however, needs a correction. Each initial nitrogen-14 nucleus receives two alpha particles, including eight additional baryons, before becoming neon-22; the intervening <beta-plus decay> also emits a <neutrino>. Thus, if the entire original metallicity is nitrogen-14 with <mass> fraction $Z_{\mathrm{initial}}$, conservation of the number of seed nuclei gives
$$
X_{22}=\frac{22}{14}Z_{\mathrm{initial}},\qquad
\boxed{\frac{\langle Z\rangle}{\langle N\rangle}
=\frac{7-Z_{\mathrm{initial}}}{7+Z_{\mathrm{initial}}}.}
$$
This is <neon-22 neutron excess from nitrogen-14>. The two expressions are compatible only when their arguments are distinguished: the displayed $(11-Z)/(11+Z)$ law uses the final <neon> <mass> fraction, not literally the earlier <nitrogen> <mass> fraction. Treating those fractions as equal is an extra simplifying approximation, not a consequence of the stated capture chain.
A <radioactively powered Type Ia supernova light curve> rises to a broad maximum over roughly weeks as energy deposited in expanding ejecta diffuses out, then declines over the following weeks and months. Its power comes predominantly from the <nickel-56 decay chain>, first <nickel> to <cobalt> and then <cobalt> to <iron>. Gamma rays and charged decay products deposit heat, which is reprocessed into optical and infrared radiation. At later times, declining <cobalt> decay power and increasing gamma-ray escape shape the tail. It is not sustained nuclear fusion in the expanding ejecta. The amount of radioactive <nickel>, <photon> diffusion time and deposition efficiency all matter, so an uncorrected fixed <luminosity> is only an idealization.
For the requested two-isotope yield calculation, assume the same total processed iron-group <mass>, namely $1M_\odot$ as inferred from the zero-neon reference. Treat the final freeze-out ashes as nickel-56 and iron-54, neglecting residual light particles and additional weak neutronization. If $w$ is the <nickel> <mass> fraction, conservation of electron fraction gives
$$
\frac w2+(1-w)\frac{26}{54}=\frac12-\frac f{22}.
$$
Since $26/54=1/2-1/54$, this reduces to the <metallicity dependence of nickel-56 yield>
$$
\boxed{w=1-\frac{27}{11}f.}
$$
Under the paper's intended identification $f=Z=1/50$,
$$
\boxed{M_{56}=\left(1-\frac{27}{550}\right)M_\odot
=\frac{523}{550}M_\odot\simeq0.95M_\odot\quad\text{(two significant figures)}.}
$$
If instead $Z=1/50$ denotes the earlier <nitrogen> <mass> fraction, retaining the actual capture stoichiometry gives $w=1-27Z/7=323/350$ and $M_{56}\simeq0.92M_\odot$. The same <temperature> and density do not alone fix a total burned <mass>; equality of that <mass> is also implicit in the comparison. The two-isotope assumption is a bulk-ash approximation, not a complete equilibrium network with all free particles identically absent.
For otherwise comparable explosions, approximate the peak <luminosity> by a quantity proportional to the radioactive <nickel> <mass>. Let $d$ be the true distance to an extremely low-neon event, and use the solar calibration $L_\odot/L_0=523/550$. Its measured flux is $F=L_0/(4\pi d^2)$, but the inferred standard-candle distance is
$$
\boxed{\frac{d_{\mathrm{est}}}{d}
=\sqrt{\frac{L_\odot}{L_0}}=\sqrt{\frac{523}{550}}\simeq0.975.}
$$
The source is intrinsically brighter than the calibration, so the distance is underestimated by about $2.5\%$, conventionally described here as about $3\%$. Equivalently, $d/d_{\mathrm{est}}=\sqrt{550/523}\simeq1.0255$, consistent with the supplied rounded factor $1.03$. Using the literal <nitrogen> metallicity instead gives about a $4\%$ underestimate. These are conditional composition effects in this simplified model; real Type Ia distance estimation also standardizes light-curve shape and color.
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