Solution:

a. The number $\mathcal{A} = 12131415161718$ is correct. Since a correct description of the number does not depend on the order, by splitting $\mathcal{A}$ into blocks of two consecutive digits

| 12 | 13 | 14 | 15 | 16 | 17 | 18 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |

and permuting them among themselves we obtain different correct numbers. The number of permutations of seven distinct blocks is
$$7! = 5040 > 2019$$

b. Every correct number is made up of two-digit blocks $ab$ whose first digit describes the second. A digit that is described in some block can appear at most $9$ times in the whole number: otherwise it could not have been correctly described in the block in question. This observation immediately limits the number of blocks to nine per digit, that is $9 \cdot 10 = 90$, hence the number of digits to $90 \cdot 2 = 180$.
The integers with at most $180$ digits (and, consequently, the correct ones) are finite in number.

c. Let us refine the upper estimate from the previous point: if a natural number has more than $91$ digits then, by the pigeonhole principle, there is a digit that appears at least $10$ times and that therefore can never be described. Observing that the number of blocks in a correct number is at most nine times the number of digits that are described, we obtain an upper bound of $9 \cdot 9 = 81$ blocks, which corresponds to a bound of $81 \cdot 2 = 162$ digits.

We now expect that the largest correct number is
$$\mathcal{G} = \underbrace{9898 \ldots 98}_{9\ \text{blocks}}\ \underbrace{9797 \ldots 97}_{9\ \text{blocks}}\ \ldots\ \underbrace{9090 \ldots 90}_{9\ \text{blocks}},$$
but it will not be necessary to prove this. Once it has been observed that it has $9 \cdot 9 \cdot 2 = 162$ digits, the proof is concluded: even if there existed a correct number larger than $\mathcal{G}$, it would still have $162$ digits.