 Solution

At a separation of $1$ fm, the proton Coulomb energy is approximately $V=e^2/(4\pi {\epsilon }_{0}r)\simeq 1.44$ MeV, whereas $k_{B}T\simeq 0.862$ 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 $\mu =m_p/2$. ThensoTreating 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 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 $1.7\times 10^3$ fm, not one femtometre.