3-digit number divisible by 6 are: 102, 108, 114,... , 996

This is an A.P. in which $ a $ = 102, $ d $ = 6 and $ l $ = 996

Let the number of terms be $ n $. Then $ t $_n = 996.

$\therefore a $ + $\left(n - 1\right)$d = 996

$\Rightarrow$ 102 + $\left( n - 1\right)$ x 6 = 996

$\Rightarrow$ 6 x $\left( n - 1\right)$ = 894

$\Rightarrow$ ( n - 1) = 149

$\Rightarrow n $ = 150

$\therefore$ Number of terms = 150.