Let $\underline{v}:\mathbb{R} ^3 \rightarrow \mathbb{R}^3$ by $\underline{v}(\underline{x})=\underline{x},\ \underline{x}=(x,y,z).$
Let $ \Omega \subset \mathbb{R}^3\ $ be a region for which $||x|| \leqslant R\ \forall \underline{x} \in \Omega\ \cup\ (\partial\Omega).$
Use the divergence theorem to prove that $Vol(\Omega)\leqslant \frac{R}{3}Area(\partial\Omega).$
What shape must $\Omega$ be?
This very much looks like I could use Cauchy-Schwarz inequality, although I am having issues formulating it.
Your vector field is $v(\vec{x})=(x,y,z)$. The divergence theorem tells you that $$ \int\int\int_{\Omega}\nabla \cdot \vec{v}\mathrm dv=\int\int_{\partial \omega}\vec{v}\cdot \vec{n}\mathrm ds $$ But the divergence of $\vec{x}=\vec{v}(\vec{x})$ is just $3$. Thus, the above expression is just $$ 3\int\int\int_{\Omega}\mathrm dv=3\text{Vol}(\Omega)=\int\int_{\partial \omega}v\cdot \vec{n}\mathrm ds $$ Slapping on some absolute values $$ 3\text{Vol}(\Omega)=\bigg|\int\int_{\partial \omega}v\cdot \vec{n}\mathrm ds\bigg|\\ \stackrel{\text{triangle ineq.}}{\leq} \int\int_{\partial \omega}|v\cdot \vec{n}\mathrm |ds\\ \stackrel{||v||=||x||\leq R}{\leq} R\text{Area}(\partial \Omega) $$ since $v(\vec{x})=\vec{x}$ yields that $||v||\leq R$.
As for what shape your $\Omega$, assuming what is intended is that there are vectors in $\Omega$ with norms all values between $0$ and $R$, you have a ball of radius $R$. Recall also the formula for the volume of a ball and it's surface area.