This is a comment, but it is a bit too long, so I will post it as an answer. At best, this is the restatement of the obvious proof strategy.
Let $A$ be an annulus, and let $\Gamma_1$ be the family of curves homotopic to the generator of the fundamental group of the annulus.
From the definition,
$$Mod(A)^{-1} = {\rm Ext}(\Gamma_1) = \sup_{\rho} \sup_{\gamma inf_{\gamma \in \Gamma_1} \frac{\ell_\rho(\gamma)^2}{{\rm Area}_\rho(A)}$$,
where the supremum is over the metrics in the conformal class of $A$. This means that if you do the calculation for a specific metric $\rho$, you get a lower bound for ${\rm Ext}(\Gamma_1)$ which gives you an upper bound for ${\rm Mod(A)}$. The picture you draw is suggestive for the possible choices of $\rho$.
To get lower bounds, one way is to use the identity
$$Mod(A) = {\rm Ext}(\Gamma_2) = \sup_{\rho} \sup_{\gamma inf_{\gamma \in \Gamma_2} \frac{\ell_\rho(\gamma)^2}{{\rm Area}_\rho(A)}$$,
where $\Gamma_2$ is the family of curves connecting the boundaries of the annulus. Now doing the calculation for a specific $\rho$ gives you a lower bound for ${\rm Mod(A)}$.
Perhaps by choosing the metrics to some interpolated versions of the flat metrics on the rectangles you can get reasonable upper and lower bounds.
