The correct option is B 200
5×(3,13,23,33,43,53,63,73,83,93)+5×(30,31,32,33,34,35,36,37,38,39)+(301,302,303,304,...,399)
=5×10+5×10+100=200
Explanation: In every set of 100 numbers there are 10 numbers whose unit digit is 3. Similarly in every set of 100 numbers there are 10 numbers whose tens digit is 3 and there are total 100 numbers whose hundreds digit is 3.