Answer
$$(e-1)\left(1-e^{-2}\right)$$
Work Step by Step
Given $$ f(x, y, z)=x e^{y-2 z} ; 0 \leq x \leq 2, \quad 0 \leq y \leq 1, \quad 0 \leq z \leq 1$$
Since we can write $f(x,y,z) =f_1(x)f_2(y)f_3(z)$, then
\begin{aligned}
\iiint_{\mathcal{B}} f(x,y,z)d V &= \iiint_{\mathcal{B}} x e^{y-2 z} d V \\
&=\int_{0}^{2} \int_{0}^{1} \int_{0}^{1} x e^{y-2 z} d z d y d x\\
&= \int_{0}^{2} \int_{0}^{1} \int_{0}^{1} x e^{y}e^{-2 z} d z d y d x\\
&=\left(\int_{0}^{2} x d x\right)\left(\int_{0}^{1} e^{y} d y\right)\left(\int_{0}^{1} e^{-2 z} d z\right) \\
&= \left(\frac{1}{2}x^2\bigg|_{0}^{2} \right) \left(e^{y}\bigg|_{0}^{1}\right) \left(\frac{-1}{2}e^{-2 z} \bigg|_{0}^{2}\right) \\
&= (e-1)\left(1-e^{-2}\right)
\end{aligned}
