= Solution
The radiation registered in the stated <solar luminosity> includes two <Positron> annihilations but excludes the two escaping <neutrinos>. Thus the visible energy per completed helium-producing chain is
$$
\boxed{Q_{\rm vis}=24.68+2(1.02)-2(0.26)=26.20\ \mathrm{MeV}.}
$$
The signs in this energy ledger matter: adding the <neutrino> energy would overestimate the energy available to sustain the measured radiation. Using $1\ \mathrm{MeV}=1.602176634\times10^{-13}$ J, the chain rate and hydrogen-consumption rate are
$$
\dot N_{\rm chain}=\frac{3.86\times10^{26}}{26.20(1.602176634\times10^{-13})}
=9.196\times10^{37}\ \mathrm{s^{-1}},\qquad
\dot N_H=4\dot N_{\rm chain}=3.678\times10^{38}\ \mathrm{s^{-1}}.
$$
With a year taken as $365.25$ days, the stated elapsed time is $1.4516496\times10^{17}$ s. Consequently,
$$
\boxed{N_{H,\rm used}=5.34\times10^{55},\qquad
N_{H,\rm remaining}=8.47\times10^{56}.}
$$
At the same fixed <luminosity>, the remaining time would be
$$
\boxed{t_{\rm remaining}=\frac{9\times10^{56}-5.3395\times10^{55}}{3.6782\times10^{38}}
=2.302\times10^{18}\ \mathrm{s}=7.29\times10^{10}\ \mathrm{yr}.}
$$
Only about $5.93\%$ of the stipulated initial reservoir has been consumed under these assumptions. This is a calculation for an entirely accessible hydrogen reservoir with constant <luminosity>. It is not a prediction of the Sun's remaining <main-sequence lifetime>: only part of the hydrogen participates in core burning, and stellar structure and <luminosity> evolve.
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