Find the noise settings to reproduce it
What you'd often be looking for would the variance – the expectation of the squared difference to the mean value in that region. If the noise you add to the actual image is zero-mean, then the mean of the flat region is (in expectation) the actual intensity, and subtracting it from all pixels and squaring the result would give you the noise power.

However, shot noise is not zero-mean:

each flat-zone image pixel $v_i$  follows an actual-intensity-"offset" poisson distribution 

$$v_i = m + n_i,\quad n_i\sim\text{Pois}(\lambda)$$

with $m$ being the actual intensity of the object, $n_i$ being shot noise samples, and $\lambda$ the intensity of the Poisson variable.

Now, since shot noise and actual image are independent,

$$\mathbb E(v_i) = \mathbb E(m + n_i) = \mathbb E(m) +  \mathbb E(n_i) =   m + \lambda\text.$$

Sadly, since we don't know the actual $m$  a priori, we'll have to dig one moment  deeper:

\begin{align}
\DeclareMathOperator{\Var}{Var} \Var (v_i) &amp;= \mathbb E\left((m + n_i- \mathbb E(v_i))^2\right)\\
&amp; = \mathbb E\left((m + n_i- m - \lambda)^2\right) \\
&amp; = \mathbb E\left(( n_i - \lambda)^2\right) \\
&amp;=: \Var(n_i)\\
&amp;=\lambda
\end{align}

So, with the Variance of your observation sample as estimator for the $\lambda$ of your Poisson variable, you get the sole statistical property of your shot noise for free. If you want to know the actual flat color, you'd also subtract that from the average of the flat region.