I'm sure I've made a trivial error but I cannot spot it.
Fix R>0 Consider the cube $C_R$ as the cube from (0,0,0) to (R,R,R) (save me from listing the 8 vertices)
Consider $S_R$ as the surface of $C_R$
Consider the vector field $v:\mathbb{R}^3\rightarrow\mathbb{R}^3$ given by $v(x,y,z) = (3x+z^2,2y,R-z)$
Part 1
Calculate $\nabla.v$
$\nabla.v=\frac{\partial{v_x}}{\partial{x}}+\frac{\partial{v_y}}{\partial{y}}+\frac{\partial{v_z}}{\partial{z}}=3+2-1=4$
Part 2
Calculate $\iiint_{C_R}\nabla.vdV$
$=\int_0^R\int_0^R\int_0^R4dxdydz$
This is trivial it is $=4R^3$
Part 3
Calculate the flux $\iint_{S_R}v.ndA$ where n is the unit normal to $S_R$ at the point.
Every keystroke is lagging now.
I did this by doing it over all 6 sides of the cube. The normals are trivial and the sides look like $R(1,s,t)$ for $s,t\in[0,1]$ (this is the right side) or $R(0,s,t)$ which is the left side.
I'm not even sure how you'd get an $R^3$ in there, I can provide more working if needed but it really isn't hard.
What have I done? Is this result right and perhaps I have misunderstood something?
Addendum
I think my error may come from my parameter ranges, I should be going from 0 to R not 0 to 1. This is essentially a substitution where I didn't multiply by the rate of that substitution with respect to the thing it replaces.
Hint
You can do $$\iint_{S_R} v\cdot n\ dA = \int_0^R \int_0^R v(0,x',y') \cdot -e_1 + v(R, x', y') \cdot e_1 + v(x', 0, y')\cdot-e_s + \ldots dx' dy'$$ The double integral will yield a $R^2$-like term and the integrand should give another (By Gauss, the results of (1) and (2) should be equal).
$$\begin{align*} v(0,s,t) \cdot -e_1 & = -t^2 \\ v(R,s,t) \cdot +e_1 & = 3R + t^2 \\ v(s,0,t) \cdot -e_2 & = 0 \\ v(s,R,t) \cdot +e_2 & = 2R \\ v(s,t,0) \cdot -e_3 & = -R \\ v(s,t,R) \cdot +e_3 & = 0 \\ \Rightarrow \iint_{S_R} v\cdot n\ dA & = \iint_{(0,0)}^{(R,R)} 4R\ ds\ dt = 4R^3 \end{align*}$$