= Solution
At a separation of $1$ fm, the <proton> Coulomb <energy> is approximately $V=e^2/(4\pi\varepsilon_0r)\simeq1.44$ MeV, whereas $k_BT\simeq0.862$ keV at the stated solar-core <temperature>. As a deliberately crude rectangular-barrier estimate, take a width of $1$ fm and use the two-proton <reduced mass> $\mu=m_p/2$. Then
$$
2\alpha L\simeq\frac{2(1\,\mathrm{fm})\sqrt{2(469\,\mathrm{MeV}/c^2)(1.44\,\mathrm{MeV})}}{197.3\,\mathrm{MeV\,fm}/c}
\simeq0.37,
$$
so
$$
\boxed{P_{\rm rectangular}\sim0.69.}
$$
Treating one <proton> as fixed and using $m_p$ instead gives about $0.59$. Both estimates only say that this arbitrarily shortened barrier would be fairly transparent; the opaque-barrier exponential is itself uncontrolled for such a small exponent.
There is an important conditioning issue. If the <protons> have \b[already] reached a separation of one femtometre, most of the long-range Coulomb suppression has already been overcome. The rectangular estimate cannot represent the probability for thermal <protons> starting far apart. Their classical turning separation at this <energy> is about $1.7\times10^3$ fm, not one femtometre.
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