2048-bit RSA Decryption
So if $M \approx 10^{20}$, and you have $4 * 10^{20}$ operations, why not just bruteforce it?

For more efficient solution, consider meet-in-the-middle technique. For all $1 \le a \le 4\sqrt{M} \approx 10^{20}$, make a hash table with values $a^e \mod{N}$. Then for all $1 \le b \le 4\sqrt{M} \approx 10^{20}$ check if $M^e/b^e \mod{N}$ is in the table.

Indeed, $M^e / b^e \equiv (ab)^e / b^e \equiv a^e \pmod{N}$ so it should be in the table (of course b must be invertible, otherwise it contains a factor of N and you can simply factorize N).

So total number of operations is $\approx 8\sqrt{M}$.