To set the stage, let me recall the well-known isodiametric inequality, which states that $$ \mathcal L^n(A)\le\frac{\alpha(n)}{2^n}(\operatorname{diam} A)^n $$ for every set $A\subseteq\mathbb R^n$, where $\alpha(n)$ is the volume of the unit ball in $\mathbb R^n$.
Clearly, the converse inequality does not hold, since one can easily figure out a sequence of sets with constant Lebesgue measure whose diameters blow up. On the other hand, if one considers the "inner diameter" of $A$: $$ \underline{\operatorname{diam}}A:=2\sup_{x\in A}\operatorname{dist}(x,\partial A), $$ then one should be able to prove the following "inner isodiametric inequality": $$ (*)\quad\quad \frac{\alpha(n)}{2^n}(\underline{\operatorname{diam}} A)^n\le \mathcal L^n(A). $$ In fact, the left-hand side of (*) is the volume of the largest ball included in $A$.
Now my question. I have a set $\Omega\subseteq\mathbb R^n$. Given $A\subseteq \Omega$, I define the relative inner diameter of $A$ with respect to $\Omega$ as follows: $$ \underline{\operatorname{diam}}_{\Omega} A:=2\sup_{x\in A}\operatorname{dist}(x,\overline{\Omega\setminus A}). $$ What I am after is a bound similar to (*), but with $\underline{\operatorname{dist}} A$ replaced by $\underline{\operatorname{dist}}_\Omega A$. Intuition suggests that there exists some $\delta_\Omega$ such that the following implication holds true: $$ \mathcal L^n(A)\le\delta_\Omega\quad\Rightarrow\quad\frac{\alpha(n)}{2^n}(\underline{\operatorname{diam}}_{\Omega} A)^n\le \mathcal L^n(A). $$ However, I am quite sure that one needs to ask something on $\Omega$ (if $\Omega$ has a cusp then I can easily imagine a counterexample). Perhaps the segment property (as in Adams' book) will do?
Thanks in advance for your help.