The first $9$ pages contain $1$ digit each.
The next $90$ pages contain $2$ digits each.
Suppose the book has $n$ pages containing $3$ digits each.
$\therefore\enspace\ 9\times 1+90\times 2+3n=$
$\ 408$
$\therefore\enspace\ 189+3n=$
$\ 408$
$\therefore\enspace\ 3n=$
$\ 219$
$\therefore\enspace\ n=$
$\ 73$
The book has $9+90+73=172$ pages.