= Solution
Assume a homogeneous Fe–Cr–Ni <alloy>, normal incidence, independent scattering events and the stated effective <scattering cross-sections>. Neglect absorption, other constituents and changes in neutron energy within the thin sheet. The <atomic fractions> are not supplied; \b[take $x_{\rm Fe}=0.70$, $x_{\rm Cr}=0.20$, $x_{\rm Ni}=0.10$ as an explicit representative assumption], not a composition measured or specified in the paper.
The conventional cell of a <face-centered cubic lattice> contains
$$
8\left(\frac18\right)+6\left(\frac12\right)=4
$$
atoms. The <number density> is therefore
$$
n=\frac4{a^3}=\frac4{(0.359\times10^{-9})^3}
=8.6452\times10^{28}\ \mathrm{m^{-3}}.
$$
For a mixture, the scattering probability per unit path length is $\Sigma_s=\sum_j n_j\sigma_j=n\bar\sigma_s$, where
$$
\begin{aligned}
\bar\sigma_s&=\sum_jx_j\sigma_j\\
&=[0.70(3.2)+0.20(2.8)+0.10(4.1)]\times10^{-28}\\
&=3.21\times10^{-28}\ \mathrm{m^2}.
\end{aligned}
$$
Thus $\Sigma_s=27.751\ \mathrm{m^{-1}}$. For thickness $t=10^{-3}\,\mathrm m$, the unscattered fraction $U$ satisfies $dU/dz=-\Sigma_sU$ with $U(0)=1$. The fraction undergoing at least one <neutron scattering> event is
$$
\boxed{P_s=1-e^{-\Sigma_st}=1-e^{-0.027751}=0.027370\simeq2.74\%.}
$$
The thin-sheet approximation gives $P_s\simeq\Sigma_st=2.78\%$, consistent to the accuracy of the assumed composition. A simple iron-dominated estimate $\bar\sigma_s\simeq3.2$ <barns> gives essentially the same \b[about $3\%$ scattered] conclusion.
For any other assumed composition, use the displayed weighted formula; mass fractions must first be converted to <atomic fractions>. Since the supplied elemental cross-sections span $2.8$–$4.1$ <barns>, any mixture of these three elements at the stated <number density> gives approximately $2.39$–$3.48\%$. This is a bound within the idealized mixture model, not a claim that every such composition is stable <austenite>.
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