Sun light takes 1,000/30,000/100,000/170,000/1,000,000 years bouncing around inside to then reach the Earth When light (photon particle) is generated inside the Sun, it takes a long time to bounce around inside to later escape and travel outwards.
Neutrinos escape immediately.
The numbers for the years trapped inside varies so much. I have included some numbers in the title of this question.
Here is a selection of online articles which I found to highlight the numbers:

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*https://www.wwu.edu/astro101/a101_sun.shtml :


Energy produced in the form of light keeps bouncing around inside the Sun, as though the Sun were made entirely of mirrors. A particle of light can take more than 30,000 years to reach the surface and escape!


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*https://futurism.com/photons-million-year-journey-center-sun :


It can take anywhere from a few thousand to a few million years for one photon to escape.


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*https://www.symmetrymagazine.org/article/august-2014/first-measurement-of-suns-real-time-energy :


Borexino scientists found measurements using solar neutrinos matched previous measurements using photons, revealing that the sun releases as much energy today as it did 100,000 years ago.


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*https://en.wikipedia.org/wiki/Sun :


... Process converts 4 million tons of matter into energy. This energy, which can take between 10,000 and 170,000 years to escape the core, is the Source ...


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*https://en.wikipedia.org/wiki/Sun#Sunlight_and_neutrinos :


Estimates of the photon travel time range between 10,000 and 170,000 years.


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*https://en.wikipedia.org/wiki/Borexino :


... solar activity has been consistently stable on a 100,000-year scale....


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*Why so much variation in the estimates?

The Borexino measurements directly use this number to compare/estimate what the sunlight would have been X years ago, but the specific number used is 100,000.
It is one thing to report widely ranging numbers, it is another to use a specific number in calculations to make conclusions.
Like-wise, there are calculations on when the Sun will expand or when the light will dim out, again using some specific number given here.


*In this case, why do scientists not calculate with lower bounds and upper bounds, choosing to instead take a specific number?

 A: Actually, the entire idea about a photon being emitted and then bouncing around until it reaches the surface, is hilarious: Each nuclear reaction inside the sun's core produces energies in the range of several $MeV$, both in kinetic energy and gamma rays. The photons that emerge from the surface have their wavelength maximum around $500nm$, which corresponds to energies around $2.4eV$. As such, for each gamma ray produced inside the core, millions of photons are emitted from the sun's surface.
Where did all those extra photons emerge from? There is no such thing as a correspondence between those gamma rays and surface photons, other than that they are part of the unceasing transformation of radiative energy into kinetic energy and back.
A: The idea that an identifiable photon "bounces around" inside the Sun and emerges some time later is incorrect.
The photons that are produced in nuclear fusion reactions or indeed by other emission processes (bremsstrahlung etc.) at temperatures of $1.5\times 10^{7}$ K in the solar core are gamma rays and X-rays. Obviously, the radiation we receive from the solar photosphere is mainly in the form of visible and near-infrared light.
What in fact happens is that individual photons are emitted and then absorbed on length scales that can be as short as 1 mm inside the Sun. The exact mean free path of a photon depends on the temperature and density conditions and varies as a function of radius inside the Sun.
The "100,000 year" figure is what you get if you assume indeed that the absorption of a photon is immediately followed by the re-emission of a (different) photon in a random direction - a so-called "random walk process". It is easy enough to show that the time taken for such a random walk to emerge at the solar surface is
$$ \tau = \frac{R^2}{lc}\ ,$$
where $R$ is the solar radius and $l$ is the mean free path of a photon. If you reverse-engineer this equation you will see that $l=0.5$ mm corresponds to $\tau= 10^5$ years.
On average, a photon that is emitted from a larger radius within the Sun will have a slightly lower energy, because of the temperature gradient. This effectively transfers energy from the inside to the outside and is called radiative diffusion.
The reason that there is variation in the "100,000 year" number is because of different assumptions about what to use for the average mean free path of a photon; which as I say, varies considerably with depth inside the Sun - from $<0.1$ mm in the core to 2-3 mm further out. It is also because it is not correct to say that each photon travels a set distance $l$ and is then absorbed. In reality there is a distribution of distances, which result in additional numerical factors in this back-of-the-envelope calculation. There is also the issue that convective heat transport rather than radiative diffusion becomes the more dominant heat transfer process in the outer 20% (by radius) of the Sun.
However, it is approximately correct to say that the neutrino flux we see today will be reflected in the photospheric luminosity of the Sun in about $10^5$ years, where I am deliberately using one significant figure. Even this is debatable. I think this would represent the timescale on which the photospheric luminosity would begin to change. For the star to reach some new equilibrium luminosity would take a longer Kelvin-Helmholtz time.
