Let’ assume we have a hash function with an output space of $2^b$. Then, the number of hash computations required to find a collision with probability 0.5 is roughly $2^{b/2}$.
Proof: From a past post we know that the probability of a hash collision when computing the hash value for $n$ different inputs is roughly
$$
p \approx 1-e^{\frac{-n(n-1)}{2m}}
$$
with $m=2^b$.
For large $n$ we can simplify this to
$$
p \approx 1-e^{\frac{-n^2}{2m}}
$$
Now, we can solve this equation for n:
$$
e^{\frac{-n^2}{2m}} \approx 1-p
$$
$$
\frac{-n^2}{2m} \approx \ln (1-p)
$$
$$
\frac{n^2}{2m} \approx -\ln (1-p) = \ln\left(\frac{1}{1-p}\right)
$$
$$
n \approx \sqrt{2m\ln\left(\frac{1}{1-p}\right)}
$$
For $p=0.5$ we finally get $n \approx 1.177\sqrt{m} \approx \sqrt{m} = \sqrt{2^b} = 2^{b/2}$.