= Solution
The cube is $1\leq x\leq2$, $0\leq y,z\leq1$, so $|x|=x$ throughout it. Evaluate the six outward <surface integrals> directly.
On $x=2$ the normal component is $2$, giving <flux> $2$; on $x=1$ it is $-1$, giving <flux> $-1$. On $y=1$ the normal component is $2x$, giving $\int_1^2\int_0^12x\,dz\,dx=3$; on $y=0$ it is zero. On $z=1$ the normal component is $e^{-1}$ and on $z=0$ it is $-1$, so those unit-area faces give $e^{-1}$ and $-1$.
Adding all six contributions yields
$$
\boxed{\Phi=2-1+3+0+e^{-1}-1=3+e^{-1}}.
$$
Only direct <surface integral> was used, as requested.
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