Square Numbers' Sequence

1-digit: $1^2$ , $2^2$ , $3^2$ = $3$ digits

2-digits: $4^2$ , $5^2$ , ... $9^2$ = $6 \times 2=12$ digits

3-digits: $10^2$ , $11^2$ , ... $31^2$ = $22 \times 3=66$ digits

$66+12+3=81$ digits, $100-81=19$ digits

19th digit after $31^2$ is the 3rd digit of $36^2=1296$

The final answer is 9.

1
2
3
4

string=''
for i in range(1,100):
    string+=str(i**2)
print string[99]

Python code, the cheap way ;)