Is direct RSA encryption of AES keys secure? I am wondering: If we take this scheme/procedure and each of it seems very secure (to me at least), is it truly secure or is there a vulnerability hidden in the process?

This is the scheme:

Bob has an RSA key with modulo $N$ with a size that is considered safe, 2048 and a public power of $e=3$ (should assure efficient encryption). 

Alice wants to send Bob a big file, and chooses symmetric encryption: She uses a random $k$ for AES and sends it encrypted using RSA using $C=k^e \bmod N$, and then sends the file encrypted by AES using key $k$.

To decrypt the file, Bob recovers $k$ using $k=C^d \bmod N$ and then decrypts the encrypted file using AES with $k$ is the key.

Is this procedure really secure?

On the paper, it uses secure parameters and seems secure, but I am not sure because $k$ is used too much here. Is there some hidden vulnerability I am missing here?

EDIT: what i am asking is in regards to attacking it, so could you please put an emphasize on attacking it rather than suggesting an alternative? i don't fully understand it, i understand that because of AES, $k^3$ cannot be more than 768 bits, so it does not pass the modulo (that is 2048). but i don't understand the technical details very well and would appreciate if you could elaborate on it instead of on possible mitigations.

thank you very much
This scheme suffers from a classic problem of textbook RSA which is mitigated e.g. by RSA-KEM (as outlined by kelalaka) or RSA-OAEP.

When you compute $k^3\bmod N$, you'll experience that $$c=k^3\stackrel{k&lt; 2^{256}}{\leq}\left(2^{256}\right)^3=2^{768}\ll 2^{2000}&lt;N$$

Now remember how $x\bmod N$ works: If $x\geq N$, then you recursively compute and return $(x-N)\bmod N$ and else you return $x$.

Given that $c\bmod N$ fits the "else" case, $c$ is simply returned. Now an adversary can just compute $\sqrt[3]c$ as your calculator does for real numbers (with a bit more precision), you get $k$ back.

Note that the above weakness doesn't violate the RSA assumption because the assumption explicitly states that $x$ is uniformly randomly sampled from $\mathbb Z^*_N$ 1 in $c=x^e\bmod N$.

1: $\mathbb Z^*_N$ is the set $\{1,\ldots, N-1\}$ without any $x$ such that $\gcd$$(x,N)&gt;1$