Answer
$$1$$
Work Step by Step
$$I=\int_{1}^{e} \int_{1}^{x} \int_{0}^{z} (\dfrac{2y}{z^3}) \ dy \ dz \ dx \\=\int_{1}^{e} \int_{1}^{x} z^{-1} \ dz \ dx \\=\int_{1}^{e} [\ln (x)] \ dx \\=[ x \ln x -x]_1^e \\=(e-1) \times (\ln e -\ln 1) - (e-1) \\=1-0 \\=1$$
