The correct answer is $\boxed{3i – 4j}$.
The curl of a vector field is a vector field that describes the circulation of a fluid around a point. It is calculated using the following formula:
$$\text{curl}(F) = \det \begin{pmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \ F_x & F_y & F_z \end{pmatrix}$$
In this case, we have the vector field $F(x, y, z) = 2x^2i + 3z^2j + y^3k$. Substituting this into the formula for the curl, we get:
$$\text{curl}(F) = \det \begin{pmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \ 2x^2 & 0 & 3z^2 \end{pmatrix} = 6x\hat{\jmath} – 4y\hat{k}$$
Therefore, the curl of $F$ at $x = y = z = 1$ is $3i – 4j$.
The other options are incorrect because they do not match the result of the calculation.