How long does a positron last in the Sun's core?

The elementary approach would be to consider the mean free path. This is given by the number of particles per volume $n~=~N/V$ times the cross section $$ x~=~\frac{1}{\sqrt{2}n\sigma}. $$ The $\sqrt{2}$ comes from the Boltzmann distribution. The natural gas law $pV~=~NkT$ and $\sigma~=~2\pi r^2$ lets this be $$ x~=~\frac{kT}{\sqrt{2}\pi pr^2}. $$ The temperature in the sun is $p~=~1.6\times 10^7K$, the classical radius of the electron is $r~=~2.8\times 10^{-15}m$ and the pressure at the center of the sun is estimated from the average density, $1.4g/cm^3$ and its radius $6.96\times 10^{5}km$ as $p~=~9.7\times 10^{10}g/cm^2$ or $9.7\times 10^{11}Pa$. In the above formula that gives a surprising $6.5m$ for the mean free path. This might be as far as a positron travels before scattering off of either a nucleon or another electron or positron. If you assume that the positron is moving at a significant percentage of the speed of light this means it will be on this path for only $2.2\times 10^{-8}s$. If about half of these scattering events are with an electron the lifetime of a positron is then about half this number.

Interesting question. In condensed matter at room temperature, there seem to be three components to the positron lifetime: fastest is direct annihilation on free electrons (more important in metals than in insulators); then annihilation of para-positronium to two photons; then annihilation of ortho-positronium to three photons. My feeling is that you ought to be able to estimate the the direct annihilation lifetime from the number density of electrons in the Sun's core. If positronium has a hydrogen-like excitation spectrum, though, it may be that positronium atoms are unbound in the Sun's core due to the high ambient temperature and only the free-capture lifetime matters.

(This is more of a comment than an answer, but it got too long for the comment box.)