Actually, if Eve does have access to Alice's plaintext messages (and Alice doesn't do random padding as she should), then yes, Eve does have a chance, even with only two messages.
Suppose she guesses $e$ (the public exponent), and that it's not a not huge value (say, the common value 65537).
Then, for every plaintext/ciphertext pair $P_i$, $C_i$, she knows that (if $N$ is the unknown modulus):
$P_i ^ e = C_i \mod N$
or, in other words:
$P_i ^ e - C_i = k_iN$
for some integer $k_i$.
So, if she takes two plaintext/ciphertext pairs, and computes the value:
$gcd( P_1 ^ e - C_1, P_2 ^ e - C_2 )$
that is extremely likely to be the value $N$, possibly multiplied by a small integer. Now, this gcd operation might be on operands of possibly a few hundreds of millions of bits long; not something you want to do every day, but it certainly in the realm of possibility if you're a bit patient.
Now, if Alice does do random padding before doing her RSA encryption, well, Eve doesn't have access to the values $P_i$, so this observation doesn't help her.