NOTE: You have an issue with the quote tags in your post you might want to go back and fix.As I read your posts it appears that you almost always did 'practical' calculations on real circuits or something like that. I say that because you don't seem to be comfortable with values that have a huge number of digits that have to be correct. This is common with pure mathematics where we might need THOUSANDS of digits to be precise, or we need to generate thousands in order to check some theoretical feature.
You are largely correct. I certain come from a practical background where I learned to live by the mantra, "If it's good enough, it's good enough." However, I have no problem with the notion that there are applications where "good enough" might require (requires, as opposed to simply specifies) a huge number of sig figs in the calculations. But this wasn't one of them.
The TS reported their answer as 51.17 kΩ.
To which you responded that you got a different answer of R1=51194.117647 (without units, of course). You did acknowledge that it might be because rounding occurred at a different place.
When it was stated that this is an unjustified number of sig figs, you then proceeded to talk about how it was needed in order to catch mistakes and how posting it with an ending 8 would be wrong. Well, this isn't some theoretical geometric pure math problem where we need to explain differences between results that don't agree to eleven sig figs. We are not in that world, we are in a practical electronics world where even values reported to three sig figs are usually overkill.
My problem with his answer of 51.17 kΩ isn't that it is "wrong" to four sig figs, it's that it was reported to four sig figs in the first place because his work doesn't justify four sig figs, only three.
Look at the end of his work:

He rounded the result to three sig figs when he got, and then later used, 5.25 V. That means his answer is now limited to three sig figs (though there are some caveats when addition and subtraction are in play and, to really know, you need to do a proper propagation of errors). So his answer should have been reported as 51.2 kΩ.
If I had wanted to take him to task, I would have compared it to a result that was rounded to the same number of digits that he reported his answer to (while making sure that it was justified to at least that many sig figs), which would have been 51.19 kΩ. That is a disagreement that is barely outside what four sig figs should tolerate. A difference of 1 in that last digit is not meaningful, since there is a good chance that it represents rounding within the allowed tolerance. But a difference of two is likely also not significant because the number of sig figs reported is an integer value while the number of sig figs warranted is a continuous quantity, so the number that is justified might actually be 3.6 and not 4, in which case an discrepancy of 2 in the fourth sig fig might well be within the tolerance. In fact, when choosing the number of sig figs to report a result to, we have to make a decision based on the context. If a result is justified to 3.6 sig figs, do we report it to only three so that we are not implying a result better than what is justified even though it means losing some information, or do we report it to four to ensure that we are presenting all the information but are injecting some noise into it. There's no "right" answer that applies universally. If it really matters, then we should be putting explicit error limits on the reported result, otherwise we ask ourselves which is the lesser of two evils, make a decision, and move on.
Going back to his work, let's consider what his answer, reported to four sig figs, would have been had he rounded his intermediate result to four or, preferably, five sig figs like he should have.
To four sig figs, it would have been 5.252 V and the final answer would have been 51.20 kΩ. This is right at the edge of the tolerance for four sig figs and underscores the point that intermediate calculations must be rounded to at least the final number of sig figs and, preferably, at least one more. Using five sig figs, it would have been 5.2515 V and the result would have been 51.19 kΩ (in fact, at this point there would have been no intermediate roundoff involved at all because all calculations would have been exact up to the final one).