73rd digit precisely!

First, try squaring smaller strings of 1’s

11111 squared is 123454321

1111111 squared is 1234567654321

If you multiply one of these out the long way, you will see that the nth column from the left is the sum of n different 1’s, so the digit is n. This is true as long as n is less than the number of digits in the number to be squared. You can also count from the other end, so the nth digit counting right to left is also n.

In our example, the 73 digit is ... 73. Thanks to carryover, we can treat it as 3, but we also have to account for the digit before. The 72nd place has enough 1’s to carry 7 over into the 73rd, so the 73rd digit is 0.

This is just $(10^{2011}+10^{2010}+ \cdots + 10^2+10+1)^2 = 1 \cdot 10^{4022}+2 \cdot 10^{4021}+3 \cdot 10^{4020} + \cdots + 3 \cdot 10^{2} +2 \cdot 10 + 1$

If we zoom in on the useful terms, we see $73 \cdot 10^{72} + 72 \cdot 10^{71} = 80 \cdot 10^{72} + 2 \cdot 10^{71} = 8 \cdot 10^{73} +\boxed{0} \cdot 10^{72} + 2 \cdot 10^{71}$ thus it follows the $73rd$ digit is $0$ .