In the range of numbers from 1 to 200, how many numbers will contain 3

In the range single digits containing 3
only one - 3

So total single digits have 3 = 1

Double digits containing 3:
a) If unit digit = 3
8*1 - one option for unit digit i.e. 3 and 8 options for tens digit, for tens digit we are leaving 0 and 3 hence 8 options
b) If tens digit is 3
1*9 - one option for tens digit i.e. 3 and 9 options for tens digit, for tens digit we are leaving only 3.
c)Both unit and tens digit are 3 :
1*1 = 1 One option for units digit and tens digit i.e. 3

So total 2 digits have at least one three - 8+9+1= 18

3 digits containing 3:
1*9*1 : unit digit can only be 1 , i.e. 3. For tens digit we are leaving 3 hence 9 options. For hundred digit one option i.e. 1
1*1*9: Tens digit one options i.e. 3. Hundred's digit 1 option i.e. 1 and units digit 9 options because we are leaving out 3.
1*1*1: Units digit one option i.e. 3. Tens digit one option i.e. 3 and hundred digit one option i.e. 1 .

So among 3 digits total number of 3's = 9 + 9 +1 = 19

Hence total three's -> 1+18+19= 38

Ans- D

Hope it's clear.