If the average of the numbers is $10$, then their sum is $10\times 10=100$.
To maximize the largest number of the ten, we minimize the other nine. Since they must be distinct, positive whole numbers, we let them be $1,2,3,4,5,6,7,8,9$. Their sum is $45$.
The sum of nine of the numbers is $45$, and the sum of all ten is $100$ so the last number must be $100-45=55\rightarrow \boxed{\text{C}}$.