I'm not quite sure from whence the confusion stems, here.

The first nine pages (since the problem implies that leading zeroes will not be used, i.e. 01, 02, 03) only require one printed digit to label the page. After that, though it costs two printed digits for each page. Since we know the total number of digits (not the total number of pages), we must take into account how many digits each page uses.

If the number of digits was higher, say, 300, we would also have to take into account the fact that the hundredth page and beyond requires three digits for each page.

We're actually counting the number of pages. We try one digit (1-9), then two digit (10-99) then three digits (100-999)

This will be easier:

We have total 187 digits,

first nine pages (1-9) take us 9 digits, so it will be 9, our remain digits is 178

now, we can check if the book is more than 99 pages (use all 2-digit numbers). To print next 90 pages (10-99) we need 90 * 2 = 180 digits, but we only have 178 digits left, so the book is less than 100 pages. Continue we have 89 pages

=> total 98 pages

or easier, you can use the answer and plug back into the questions

A. 99 pages, digits = 9 + 90 * 2 = 189
B. 98 pages, digits = 9 + 89 * 2 = 187 (Bingo)

Thank you all! It makes sense now.