The correct answer is $\boxed{0}$.
The circulation of a vector field $\overrightarrow{A}$ around a closed path $\Gamma$ is given by the following integral:
$$\oint_\Gamma \overrightarrow{A} \cdot d\overrightarrow{l}$$
In this case, the vector field is $\overrightarrow{A} = xy\hat{\imath} + x^2\hat{\jmath}$ and the path is a triangle with vertices $(0,0)$, $(1,0)$, and $(1,1)$. The line integral over this path can be evaluated as follows:
$$\oint_\Gamma \overrightarrow{A} \cdot d\overrightarrow{l} = \int_0^1 xy\,dx + \int_0^1 x^2\,dy$$
The first integral can be evaluated using the substitution $u = x$, which gives:
$$\int_0^1 xy\,dx = \int_0^1 u^2\,du = \frac{1}{3}$$
The second integral can be evaluated using the substitution $v = y$, which gives:
$$\int_0^1 x^2\,dy = \int_0^1 v^2\,dv = \frac{1}{3}$$
Therefore, the total line integral is:
$$\oint_\Gamma \overrightarrow{A} \cdot d\overrightarrow{l} = \frac{1}{3} + \frac{1}{3} = 0$$
The other options are incorrect because they do not represent the correct value of the line integral.