$$\begin{aligned} & \mathrm{Mg}=\mathrm{F}_{\mathrm{B}} \Rightarrow\left(400 \times 10^{-3}\right)=10^3 \times \mathrm{V}_{\mathrm{d}} \\ & \mathrm{~V}_{\mathrm{d}}=400 \times 10^{-6} \mathrm{~m}^3 \\ & (\text { Vol. })_{\text {outside }}=\left(10 \times 10^{-2}\right)^3-400 \times 10^{-6} \\ & =600 \times 10^{-6} \mathrm{~m}^2=600 \mathrm{~cm}^3 \end{aligned}$$