Answer
$$2 \pi $$
Work Step by Step
A vector field is said to be conservative when $\int_C F \cdot dr=0$
We need to consider any closed path.
$$\int_C F\cdot dr=\int_{(1,0,0)}^{(1,0,2 \pi)} \nabla (x^2 ze^y) \space dr $$
Now, $$[(x^2 ze^y)]_{1,0,0}^{1,0,2 \pi}-[(x^2 ze^y)]_{1,0,0}^{1,0,2 \pi} =2 \pi -0 \\=2 \pi $$
