Cryptography Engineering - Design Principles and Practical Applications - Chapter 9 Generating Randomness - Section 9.4 The Generator
This section talks about Fortuna, where it uses a block cipher with a 128-bit block and a 256-bit key size. The cipher is used in CTR mode. In the CTR mode, if an attacker can access the internal state than then the security is lost ( the key and current counter value), therefore after a request, 2 new blocks are generated and used as a new key, and the old key is forgotten to eliminate exposure of the old requests.
All talk is about distinguishing the PRP from PRF in the CTR mode. In the CTR mode with a PRP, if you encrypt with one key then $2^{128}$ blocks are distinct since it is a permutation, but we don't expect this from a PRF. So the PRP-PRF switching lemma applies to distinguish the statistical deviation. After $q$ messages the security lost is proportional to $q^2\!/2^{128}$.
In the classical birthday attack, when the outputs are random we expect collisions, remember the CTR mode doesn't generate one with PRP. For a PRF with 128-bit output, approximately

Once we generate $2^{64}$ outputs there will be  $$\approx (2^{64})^2/2^{128}/2 = 2^{128 - 129} = 1/2$$ change collision. But we will not see one in the Fortuna, it uses PRP in CTR mode! Therefore this is a distinguisher.

if you only generate $2^{16}$ blocks for a single key then the probbaility is $$\approx (2^{16})^2/2^{128}/2 = 2^{32 - 129} = \dfrac{1}{2^{97}}$$ this means this distinguish probability is negligible.


Requesting $2^{97}$ blocks is the reverse of the birthday attack. Since one has $\dfrac{1}{2^{97}}$ change of seeing a collision for $2^{16}$ blocks then you need to request $2^{97}$ blocks to see a collision within a $2^{16}$-block. Therefore the cost is $2^{97}\cdot 2^{16} = 2^{113}$.
For 256-bit block sizes, the collision is no more a threat since you need to $2^{128}$ blocks to see one with 1/2 probability.
$$\approx (2^{128})^2/2^{256}/2 = 2^{256 - 257} = 1/2$$
The $2^{113}$ is continuation from the 128-bit attack cost, since $$(2^{113})^2/2^{256}/2 = \dfrac{1}{2^{32}}$$