Without knowing the details of the vector field, I can't give you a direct answer for the flux. At any rate, the general idea would be to use the divergence theorem, which states the flux of [tex]\mathbf f[/tex] over the closed surface [tex]\mathcal S[/tex] (surface of a cube in this case) is equivalent to the integral of the divergence of [tex]\mathbf f[/tex] over the interior of the surface (call it [tex]\mathcal R[/tex]):
[tex]\displaystyle\iint_{\mathcal S}\mathbf f\cdot\mathrm d\mathbf S=\iiint_{\mathcal R}(\nabla\cdot\mathbf f)\,\mathrm dV[/tex]
The latter integral is less work to compute, and hence the usefulness of the divergence theorem. Denoting the vector field by [tex]\mathbf f(x,y,z)=(f_1(x,y,z),f_2(x,y,z),f_3(x,y,z))[/tex], we have
[tex]\displaystyle\iiint_{\mathcal R}(\nabla\cdot\mathbf f)\,\mathrm dV=\int_{z=0}^{z=c}\int_{y=0}^{y=b}\int_{x=0}^{x=a}\left(\dfrac{\partial f_1}{\partial x}+\dfrac{\partial f_2}{\partial y}+\dfrac{\partial f_3}{\partial z}\right)\,\mathrm dx\,\mathrm dy\,\mathrm dz[/tex]
