Maximum Number of Blocks to be Encrypted under One Key in CBC and CTR Mode? What is the Limit on Maximum Number of Blocks to be Encrypted under One Key in CBC and CTR Mode? What happens if the limit is crossed and more number of blocks are encrypted?
CTR mode.
There's a generic birthday attack on CTR mode of a block cipher.

The idea of CTR mode is to use a PRF, or pseudorandom function family, $F_k\colon \{0,1\}^\iota \to \{0,1\}^n$ to approximate a one-time pad with a short key $k$: $$c = m \oplus (F_k(0) \mathbin\| F_k(1) \mathbin\| F_k(2) \mathbin\| \cdots \mathbin\| F_k(\ell - 1)),$$ where $\ell$ is the length in $n$-bit blocks of the message $m$ (ignoring details of partial blocks).

If instead of a PRF we have a PRP, a pseudorandom permutation family or $n$-bit block cipher, $E_k\colon \{0,1\}^n \to \{0,1\}^n$, we can use that to approximate a PRF.  But a PRP, being a family of permutations, has the property that in the sequence of blocks $E_k(0), E_k(1), \dots, E_k(\ell - 1)$, as long as $\ell &lt; 2^n$, there are no repeated blocks.  In a uniform random sequence of blocks, we would expect to see a collision when $\ell \approx \sqrt{2^n} = 2^{n/2}$, by the birthday paradox.

When $n = 128$ as in AES, this procedure—check for a repeated block—enables an attacker to distinguish a very long but not unimaginably long CTR mode pad from a very long uniform random pad with nonnegligible probability.

CBC mode.
In CBC mode, the $i^\mathit{th}$ ciphertext block is $c_i = E_k(m_i \oplus c_{i-1})$, with $c_{-1}$ being the initialization vector.  If the adversary notices that $c_i = c_j$ then they can infer $$m_i \oplus c_{i-1} = m_j \oplus c_{j-1}$$ from which they learn $$m_i \oplus m_j = c_{i-1} \oplus c_{j-1}.$$  Turning this into a distinguisher and calculating how the birthday bound is relevant here is left as an exercise for the reader who has had more breakfast so far this morning than I have.

Recommendations.
Because of these birthday attacks, cryptographers usually sternly advise against using AES-CTR or AES-CBC for more than a few exabytes of data, or, to be conservative, a few hundred terabytes.  (It doesn't matter whether that's for a lot of short messages or a few long messages.)  64-bit block ciphers like 3DES fare even worse, to the point that they can be broken in practice.

When $n = 256$ as in Threefish-256, these attacks are irrelevant: the birthday bound is around $2^{128}$ blocks of data, which you will never see.  When it's a PRF with $n = 512$, as in Salsa20 or ChaCha, it's even less relevant—although if you pick per-message nonces at random, the birthday bound once again becomes relevant because those nonces are only 64 bits and there is a nonnegligible chance of a security-destroying collision in the nonce, which is why the extended-nonce variants XSalsa20 and XChaCha exist and why some cryptographers and standardizers are considering nonce-misuse-resistant AEAD schemes.   In sum, the birthday paradox is a versatile tool!