= Solution
Use GeV consistently with the supplied $A'$:
$$
m_u=0.362\,\mathrm{GeV},\qquad m_s=0.537\,\mathrm{GeV},\qquad A'=0.026\,\mathrm{GeV^3}.
$$
The spin-independent sum is $m_u+2m_s=1.436\,\mathrm{GeV}$. The <spin-spin interaction> corrections are
$$
\begin{aligned}
\Delta M_{1/2}&=0.026\left[\frac1{4(0.537)^2}-\frac1{(0.362)(0.537)}\right]\mathrm{GeV}
=-0.111208\,\mathrm{GeV},\\
\Delta M_{3/2}&=0.026\left[\frac1{4(0.537)^2}+\frac1{2(0.362)(0.537)}\right]\mathrm{GeV}
=+0.0894151\,\mathrm{GeV}.
\end{aligned}
$$
Thus
$$
\boxed{M(\Xi^0)\simeq1324.8\,\mathrm{MeV},\qquad
M(\Xi^{*0})\simeq1525.4\,\mathrm{MeV}.}
$$
The displayed decimal places document the calculation, rather than asserting that this simple <constituent-quark model> has sub-MeV predictive precision.
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