Required numbers are 102,108,114,....996
This is an A.P in which =102,d=6, l=996
Let the number of terms be n. Then,
a+(n−1)d=996
⇒ 102+(n−1)×6=996
⇒ 6×(n−1)=894
⇒ (n−1)=149
⇒ n=150
Hence there are 150 three digits numbers which are divisible by 6.