First such number is  105.
Last  such number is  987.

How many intervals by 21 integer numbers are between these two numbers?
It is

 = 42.

So, there are 42 intervals.
Hence, the number of multiples 21 from 105 to 987 inclusively is 42 + 1 = 43.

Answer. There are 43 multiples of 21 that are 3-digit integers.

The smallest 3-digit number is 100 and the largest
3-digit number is 999.

Dividing 100 by 21 gives 4.76, which rounds up to 5,
so the smallest 3-digit multiple of 21 will be 5*21 or 105.

Dividing 999 by 21 gives 47.57, which rounds down to
47, so the largest 3-digit multiple of 21 will be 47*21 or 987.

So the 3-digit multiples of 21 form the arithmetic sequence 

5*21,6*21,...,46*21,47*21

Dividing them all by 21 gives the sequence

5,6,...,46,47, which has the same number of terms.

Subtracting 4 from each gives the sequence

1,2...,42,43, which also has the same number of terms.

Answer: 43

Edwin