The correct answer is $\boxed{\frac{1}{2}}$.
Green’s theorem states that the line integral of a vector field over a simple closed curve $C$ is equal to the double integral of the curl of the vector field over the region enclosed by $C$. In this case, the vector field is $F(x, y) = xy \hat{\imath} – y^2 \hat{\jmath}$ and the curve $C$ is the square cut from the first quadrant by the lines $x = 1$ and $y = 1$.
The curl of $F$ is $F_y = x$ and $F_x = -y$. Therefore, the double integral of the curl of $F$ over the region enclosed by $C$ is
$$\iint_R (-y) \, dx \, dy = \int_0^1 \int_0^1 -y \, dx \, dy = \int_0^1 \left[ -\frac{y^2}{2} \right]_0^1 dy = \frac{1}{2}.$$
By Green’s theorem, the line integral of $F$ over $C$ is also equal to $\frac{1}{2}$.
The other options are incorrect because they do not take into account the orientation of the curve $C$. The curve $C$ is positively oriented, which means that we are integrating in the counterclockwise direction. If the curve $C$ were negatively oriented, the line integral would be equal to $-\frac{1}{2}$.