So we are dealing with hundreds. It has three places for digits. I'm taking more or less brute force way.
_ _ _
Let's start with one place. We can force it to be six, then count how many possible digits for other places. We forbid other place to be six for now so counting would be easier.
_ _ 6
Hundred place can be 2, 3, 4, 5, or 7, so that would be 5 possible digits for hundred place
Ten place can be 0,1, 2, 3, 4, 5, 7, 8, 9 so that would be 9 possible digits
So there are 9 × 5 = 45 numbers between 200 and 800 with 6 in only one place.
Now do ten place:
_ 6 _
Again hundred place has 5 possible digits
One place would have 9 possible digits
So again there are 45 numbers between 200 to 800 with 6 only in ten places.
Now put 6 in hundred place
6 _ _
Ten and one place can be 0, 1, 2, 3, 4, 5, 7, 8, 9, so that's 9 × 9 = 81
So then we add up all amount of numbers we gathered. 45 + 45 + 81 = 171. However we under-counted because we did not allow other places to be 6.
Let's count how many numbers have exactly two 6's We can do the same thing. Let one and ten place be 6.
_ 6 6
So possible digits for hundred place would be 2,3,4,5,7, so that's 5 numbers
Now do 6 _ 6. Ten place can be 0, 1, 2, 3, 4, 5, 7, 8, 9, so that's 9 numbers.
Finally, 6 6 _. Again same as ten place, there's another 9 numbers.
So 171 + 5 + 9 + 9 = 194
Finally there's 666, so we add one then we get 195.
