There are 9 1-digit numbers, 90 2-digit numbers, 900 3-digit numbers and 9000 4-digit numbers. Hence the last digit of the last 4-digit number is in position $\displaystyle 9+2\times90+3\times900+4\times9000=38889.$
The last digit of the last 5-digit number will be in position $\displaystyle 38889+5\times90000=488889$ which is greater than 206787. Hence the number we are looking for is a 5-digit number. The position of the last digit of the 5-digit number $\displaystyle N$ is $\displaystyle 5(N-9999)+38889=5N-11106.$ Now find the largest $\displaystyle N$ such that $\displaystyle 5N-11106\le206787.$
$\displaystyle 5N-11106\,\le\,206787$
$\displaystyle \implies\ 5N\,\le\,217893$
$\displaystyle \implies\ N\,\le\,43578.6$
So that number is $\displaystyle N=43578,$ whose last digit is in position 206784. The next 5-digit number is 43579 and the 206787th digit in the sequence is the 3rd digit of this number, which is 5.