Edit: Some of my text was removed because I don't agree with it.41%
If you square the number $a+10b+100c+\ldots$ (where $a,\,b,\,c\ldots$ are digits from 0 to 9) then you getI really meant to ask (hopefully an easier question with less work):
"What percent of the perfect squares have an odd digit in their tens place?"
That is what I get.If you square the number $a+10b+100c+\ldots$ (where $a,\,b,\,c\ldots$ are digits from 0 to 9) then you get
$a^2 + 20ab + \ldots$ (everything else is a multiple of 100).
If you are only interested in whether the tens digit is even or odd, then a multiple of 20 makes no difference. So everything depends on whether the tens digit in $a^2$ is even or odd. The squares of the digits from 0 to 9 are
00, 01, 04, 09, 25, 49, 64, 81 (tens digit even)
16, 36 (tens digit odd).
Each of these is equally likely to occur, so I reckon that the proportion of perfect squares with an odd digit in their tens place is 20%.
... and I agree with your 41% for those with an odd digit in the hundreds place. I simply counted the number of odd hundreds-place digits in the list of the first 100 squares (as listed here, for example). The proportion in any other sequence of 100 consecutive squares will be the same.That is what I get.