The assumption has 2 problems with it. 1) the universe is expanding, so it's larger than what we observe and the rate of that expansion isn't exactly known. 2) we don't know how old the universe is exactly (They have a pretty darn good estimate, but it's still an estimate) and 3) due to gravity and time dilation, the observable universe isn't a perfect sphere. It's sightly lopsided and bubbled - so we can't measure it's size in perfect terms.
But lets say that right "now", the universe is exactly 13.8 billion years old and lets pretend it's a perfect sphere with 13.8 billion light years radius around us. (We'll address the expansion later).
Diameter of a hydrogen atom (source): http://en.wikipedia.org/wiki/Bohr_radius
5.2917721092(17)×10−11 M, which means, 1.88972 x 10^10 hydrogen atom diameter in 1 meter
and meters in a light-year: 9.4605284 × 10^15, and that times 13.8 x 10 ^9.
So, right "now" assuming it's the same now it was before, our observable universe is (pretending it's a perfect sphere), about 2.467 x 10^36th hydrogen atom radii in the radius. Now, if we're calculating the circumference to within the radius of a hydrogen atom, we have to multiply this by 2 Pi (cause that's what we're estimating, we know the Radius exactly), so, the circumference is about 1.55 x 10^37 or, lets use E37.
So, how many digits of Pi would you need - lets look at Pi.
The accuracy of each digit depends on the number that follows after it, for example, if we don't know the next digit, it could be a handful of 9s, so if we have the first digit (3) and we don't know the decimals, our maximum error is .99999999(etc)/3 or 33.3% - curious thing about numbers like Pi, there will always be at some point in it's infinity of numbers, every combination of consecutive 9s, for example 99, 999, 9999, 99999, 6 - 9s, 7-9s, nine hundred and ninety nine nines, even a million billion trillion 9s, all of them, at some point in it's infinite string of numbers, in a row. There has to be, otherwise it wouldn't be truly random. But I digress.
But back to the problem, while the error varies with the numbers that follow, the error generally is on average, improved by a factor of 10 with each digit.
So, we need a percentage error lower than 1/1.55 E 37. And the answer to that is fairly simple. 38 digits (I'm counting the 3 as a digit, if you only count digits as past the decimal, then you just need 37) - now, the youtube guy says 39. Maybe he's taking into account expansion. Not sure why I have different numbers than him, but we're close.
To add 2 digits to 40, it's just 100 times smaller - give or take.
Now, Planck Lengths, the answer is quite different. There are nearly as many Plank Lengths in the diameter of 1 hydrogen atom as there are hydrogen atom diameters in 1 light year. (Well, not almost as many - but close enough) - that's a pretty good way to define how small a Planck Length is - it's really tiny.
There's about 3.27 x 10^24th Planck length's in the radius of 1 hydrogen atom. So, the margin for error 1/5.08 x 10^51. so, you'd need 62 digits of Pi to measure the circumference of the universe to within one Planck length, and if we calculate for expansion (about 46 billion light years radius not 13.8), you'd need another digit, so 63.
All this is just exponent multiplication (which is basically addition) - nothing fancy.
A penny on the national debt for example is 1/1.8E15, so my Excel spreadsheet with 16 digits can count the national debt to the penny. Add just a couple more decimals and you can calculate daily interest rates on the changing national debt to fractions of pennies and fun stuff like that.
Now, if your working with public key cryptography (whether encrypting or decrypting) which uses very very big prime numbers, then you need over 100 digits (and a very very fast computer if you're decrypting), so there is at least one practical use for having over 100 digits in some calculations. Mostly it's overkill though.
Mersenne number hunters need to work with millions and millions of digits, but that's just perfect number chasing. Not much practical use there.