Solution:

The answer is $\mathbf{(C)}$. We can first exclude from the possible prime numbers all even numbers, and thus consider only sets of 10 consecutive odd numbers greater than 50. Among these, at least three are multiples of 3 and exactly two are multiples of 5. The two multiples of 5 differ by 10, so at most one of them is also a multiple of 3. It follows that there are at least $3+2-1=4$ odd numbers that are multiples of either 3 or 5, and are therefore not prime. The total number of possible primes is therefore less than or equal to 6.

The number sought is however exactly equal to 6, as shown by the example of the 6 prime numbers $97, 101, 103, 107, 109, 113$ contained in the interval $\{96, \ldots, 115\}$.