Use the divergence theorem.
[tex]\mathbf f(x,y,z)=(x^3,y^3,z^3)\implies \nabla\cdot\mathbf f=3(x^2+y^2+z^2)[/tex]
Calling the closed region [tex]\mathcal R[/tex] and its boundary the surface [tex]\partial\mathcal R[/tex], we have by the divergence theorem,
[tex]\displaystyle\underbrace{\iint_{\partial\mathcal R}\mathbf f\cdot\mathrm d\mathbf S}_{\text{flux}}=\iiint_{\mathcal R}\nabla\cdot\mathbf f\,\mathrm dV[/tex]
We set up the volume integral in cylindrical coordinates, using
[tex]\begin{cases}x=u\cos v\\y=u\sin v\\z=w\end{cases}\implies\mathrm dV=u\,\mathrm du\,\mathrm dv\,\mathrm dw[/tex]
[tex]\displaystyle\iiint_{\mathcal R}\nabla\cdot\mathbf f\,\mathrm dV=3\int_{w=0}^{w=6}\int_{v=0}^{v=2\pi}\int_{u=0}^{u=8}(u^2\cos^2v+u^2\sin^2v+w^2)u\,\mathrm du\,\mathrm dv\,\mathrm dw[/tex]
[tex]=6\pi\displaystyle\int_{w=0}^{w=6}\int_{u=0}^{u=8}(u^3+uw^2)\,\mathrm du\,\mathrm dw=50,688\pi[/tex]
