Birthday attack on AES-CCM with 32-bit Authentication Tag
Please note: This answer is entirely based on the theorem the OP stated.

Notation:


$\tau$: length of the tag, in bits, this is 32 in your case
$\sigma$: the number of encrypted / decrypted blocks
$n$: the block size of the cipher, in bits, this is 128 for AES
$q_{dec}$: the allowed number of decryption queries
$\text{Adv}$: the probability that an attacker wins "the game" and thus breaks the associated security definition


Formula: $$\text{Adv}&lt;\frac{q_{dec}}{2^\tau}+\frac{\sigma^2}{2^n}$$



First, for $q_{dec}$: You should really set this value as low as you can tolerate it to be without giving (too many) false positives, ideally this would be $1$.

The next step is to chose the upper probability you're willing to tolerate for an attacker breaking your scheme. Ideally, you'd want to set this as low as possible, for example FIPS 140-2 specifies that no random attempt should have a chance better than $\frac{1}{10^6}$ of succeeding (at wrongly authenticating the user) and it also specifies that an attacker may not have a chance greater than $\frac{1}{10^5}$ of succeeding within one minute when being alone with a cryptographic module. Also note that the chance of an attacker succeeding at least once after $k$ tries with probability $\text{Adv}$ is $1-(1-\text{Adv})^k$, so you may want to do the math and keep the value reasonably low for a given time-period and try-count an adversary has.

Now, after you figured out how much advantage you're willing to give to the attacker and how many decryption queries you're willing to accept, you can finally go ahead and use the above equation to find the maximal number of block cipher calls allowed for the scheme. This should be $$\sigma&lt;2^{48}\cdot\sqrt{2^{32}\cdot\text{Adv} - q_{dec}}$$