Explanation:

Surely 3 can occur at either hundreds place or tens place or units place. So three cases arise.

a) If 3 occurs at hundredths place then the digit at tens place can be chosen in only nine ways (all ten digits leaving only 3 so we are left with 9 digits) and digit at units place can be chosen in only 8 ways (as 3 and digit at tens place cannot be used again) So total such numbers = 1 × 9 × 8 = 72 

b) If 3 occurs at tens place then its hundreds place can be only chosen in only 8 ways (because use of 3 is not allowed and if we use 0 out of the remaining 9 digits it will be a 2-digit number which is not allowed) and unit place can be chosen only in 8 ways (since digit at hundredths place and 3 is not allowed) So total such numbers = 8 × 1 × 8 = 64

c) If 3 occurs at units place then its hundreds place can be chosen in only 8 ways (because use of 3 is not allowed and if we use 0 out of the remaining 9 digits it will be a 2-digit number which is not allowed) and tens place can be chosen only in 8 ways (since digit at hundredths place and 3 is not allowed) 

So total such numbers = 8 × 8 × 1 = 64 Hence total such numbers = 72 + 64 + 64 = 200

Hence, option B is correct.