Really stuck on this

For digits 1-9, there is 1 digit.

For digits 10-99, there are 2 digits.

For digits 100-999, there are 3 digits.

Therefore, for digits 1-9, there is a total of 9 digits.

For digits 10-99, there is a total of 2((99-10)+1) = 2(90) = 180 digits

For digits 100-999, there is a total of 3((999-100)+1) - 3(900) = 2700 digits.

There is a total of 9 + 180 + 2700 = 2889 digits.

The 2889th digit is 0, so the 2890th digit must be 1.

The 110th digit after 2890th digit is 3.

Therefore the 1000th digit is also 3.

Tell me if I'm wrong.

The 1000th digit is 3 of 370.

1 - 9    gives us  9 digits

10 - 99  gives us    [ 99 - 10 + 1] * 2  =  180 more digits

So far....we have  189 digits

So  we need     1000 - 189  =  811 more

Let us suppose we only need  810 more

Dividing this by  3  =   270

So.....we need  270  3 digit numbers  (including 100)  to give us  810digits

We can solve this equation  to find the last integer we need to write to get 810  more digits

[ n - 100 + 1] * 3  = 810

[n - 99] * 3  = 810     divide both sides by 3

n - 99  =  270      add 99 to both sides

n = 369

So  after we write 369 we will have written   9 + 180 + 810  = 999 digits

The next integer we write will be the 1000th one

This is, obviously, 3