= Solution
The gamma width is $0.10\,\mathrm{eV}=10^{-7}\,\mathrm{MeV}$, so the remaining <proton> width is $\Gamma_p=0.20-0.15-10^{-7}=0.0499999\,\mathrm{MeV}$. At the peak of the same resonance the total-width denominator and gamma exit width cancel. <alpha particles> and <carbon-12> have <spin> zero; the <proton> and the stated nitrogen level each have <spin> $1/2$. Hence $g_\alpha/g_p=4$, independently of the unspecified resonance <spin>. The <proton> <reduced mass> and <momentum> are
$$
\mu_pc^2=879.223\,\mathrm{MeV},\qquad p_p^*c=\sqrt{2(879.223)(2.113770)}=60.967\,\mathrm{MeV}.
$$
The <same-resonance entrance-channel cross-section ratio> is therefore
$$
\boxed{\frac{\sigma_\alpha}{\sigma_p}
=\frac{g_\alpha}{g_p}\frac{p_p^{*2}}{p_\alpha^{*2}}\frac{\Gamma_\alpha}{\Gamma_p}
=4\frac{60.967^2}{205.790^2}\frac{0.15}{0.0499999}\simeq1.053}.
$$
The ratio compares each reaction at its own laboratory resonance <energy>. Factors of $\hbar^2$ restoring the <cross section>'s area units cancel. Only the supplied nitrogen <spin> magnitude is used: its printed positive <parity> is immaterial to this <spin> average, so no orbital-parity inference is made from that assignment.
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