Let $\Omega\subset\mathbb{R}^3$ be a compact region whose boundary is a surface $S$. Define the radius $M:=\frac{1}{2}\max\{|x-y|:x,y\in\Omega\}$. Is it always true that $\text{Vol}(\Omega)\le\frac{1}{3}M\text{Area}(S)$? If so, can we generalize it?
Let $c\in\mathbb{R}^3$ be an arbitrary constant. Define $F(x,y,z)=(x,y,z)-c$. By divergence theorem, $$\begin{align*} \int_S F\cdot\vec{n}\,d\sigma &=\int_\Omega (\nabla\cdot F)\,dV \\ &=3\int_\Omega 1\,dV \\ &=3\text{Vol}(\Omega). \end{align*}$$ We have $$\begin{align*} \text{Vol}(\Omega) &= \frac{1}{3} \left|\int_S F\cdot\vec{n}\,d\sigma\right| \\ &\le \frac{1}{3}\int_S \left|F\cdot\vec{n}\right|\,d\sigma \\ &\le \frac{1}{3}\int_S |F||\vec{n}|\,d\sigma \end{align*}$$ If we choose $c$ to be any point in $S$, then $|F|\le 2M$. Also, $\vec{n}$ is just the unit normal, so $\int_S |\vec{n}|\,d\sigma=\text{Area}(S)$. Thus, $\text{Vol}(\Omega)\le\frac{2}{3}M\text{Area}(S)$.
However, I haven't been able to cut this bound down to the desired one, nor can I find a counterexample.