Answer
$$3\left( {e - 1} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\int_1^2 {\int_0^1 {yz{e^x}} dx} dz} dy \cr
& = \int_0^2 {\int_1^2 {\left[ {\int_0^1 {yz{e^x}} dx} \right]} dz} dy \cr
& {\text{Integrate with respect to }}x \cr
& = \int_0^2 {\int_1^2 {yz\left[ {\int_0^1 {{e^x}} dx} \right]} dz} dy \cr
& = \int_0^2 {\int_1^2 {yz\left[ {{e^x}} \right]_0^1} dz} dy \cr
& = \int_0^2 {\int_1^2 {yz\left( {e - 1} \right)} dz} dy \cr
& = \left( {e - 1} \right)\int_0^2 {\int_1^2 {yz} dz} dy \cr
& {\text{Integrate with respect to }}z \cr
& = \left( {e - 1} \right)\int_0^2 {y\left[ {\frac{{{z^2}}}{2}} \right]_1^2} dy \cr
& = \left( {e - 1} \right)\int_0^2 {y\left( {\frac{3}{2}} \right)} dy \cr
& = \frac{{3\left( {e - 1} \right)}}{2}\int_0^2 y dy \cr
& {\text{Integrate with respect to }}y \cr
& = \frac{{3\left( {e - 1} \right)}}{2}\left[ {\frac{{{y^2}}}{2}} \right]_0^2 \cr
& = \frac{{3\left( {e - 1} \right)}}{2}\left[ {\frac{{{2^2}}}{2} - 0} \right] \cr
& = 3\left( {e - 1} \right) \cr} $$
