There are coins of three different colours with one single digit number (any number from 0 to 9) printed on each side of all the coins. The numbers printed on coins of one colour cannot be repeated on coins of any other colour.

I.There are three red coins and for each red coin the difference between the numbers printed on two sides is 1 and all red coins have identical pairs of numbers.
II. There are three blue coins. The single digit numbers on six sides of these three coins are six consecutive numbers such that the sum of the two numbers on every blue coin is the same and is equal to 11.
III. There are three green coins. All three of them have the number 0 printed on one of the sides and same odd digit number on the other side.
If all nine coins are arranged such that all odd numbers are seen on the top, then what is the sum of these numbers seen on the top of nine coins?

The biggest clues we have for are for the blue coins, they are 6 consecutive numbers and the numbers on each coin add up to give 11. 
The possible pairs are (9,2),(8,3),(7,4),(6,5), of these four pairs we cannot take (9,2) because then we cannot have 6 consecutive numbers. The only way is to have the green coins as (3,8), (4,7) and (5,6) giving us the 6 consecutive numbers from 0 to 9. 

The next important clue is for the green coins, we know that one of the number is 0 and the other number is odd. we only have 1, 2 and 9 remaining. 

For the red coins, the difference between the numbers is 1 i.e, the numbers are consecutive. Out of the three remaining number the only way it is possible is if the red coins have 1 and 2 on them. So the red coins will be (1,2),(1,2) and (1,2)
Leaving only 9 for the green coins, leading the green coins to be (0,9),(0,9) and (0,9).

Now all coins are arranged with the odd sides up, so the sum of these odd sides will be {Blue coins}$$3+7+5$$ + {Red Coins}$$1+1+1$$+ {Green Coins}$$9+9+9$$ giving the net sum to be 45.

Therefore, Option B is the correct answer.