Solution:

The answer is $71$. By condition c) the product of the two digits must be a prime number. For this to happen, it is necessary that one of the two digits be $1$, otherwise the number resulting from the multiplication is composite.

Assuming that $1$ is the tens digit, condition a) restricts the possibilities to just $11$, $13$, $17$, $19$, the only prime numbers between $10$ and $19$ inclusive. Swapping the order of the digits, $11$, $31$, $71$ are prime while $91=13 \times 7$. Consequently, the largest number satisfying the three conditions a), b), c) is $71$.