Stokes' theorem says
[tex]\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\iint_{\mathcal S}(\nabla\times\mathbf f)\cdot\mathrm d\mathbf S[/tex]
where [tex]\mathcal S[/tex] is the region with boundary [tex]\mathcal C[/tex]. We can parameterize this region by
[tex]\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+(6-u\cos v)\,\mathbf k[/tex]
with [tex]0\le u\le12[/tex] and [tex]0\le v\le2\pi[/tex]. This gives us a surface element of
[tex]\mathrm d\mathbf S=(\mathbf s_u\times\mathbf s_v)\,\mathrm du\,\mathrm dv=(u\,\mathbf i+u\,\mathbf k)\,\mathrm du\,\mathrm dv[/tex]
Meanwhile, the curl of the given vector field is
[tex]\nabla\times\mathbf f=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\D_x&D_y&D_z\\xy&2z&6y\end{vmatrix}=4\,\mathbf i-x\,\mathbf k[/tex]
[tex]\equiv4\,\mathbf i-u\cos v\,\mathbf k[/tex]
So the surface integral reduces to
[tex]\displaystyle\iint_{\mathcal S}(\nabla\times\mathbf f)\cdot\mathrm d\mathbf S=\int_{v=0}^{v=2\pi}\int_{u=0}^{u=12}(4u-u^2\cos v)\,\mathrm du\,\mathrm dv=576\pi[/tex]
You can verify this by computing the equivalent line integral using the parameterization
[tex]\mathbf r(t)=12\cos t\,\mathbf i+12\sin t\,\mathbf j+(6-12\cos t)\,\mathbf k[/tex]
with [tex]0\le t\le2\pi[/tex]. You should (obviously) get the same result.
