The correct answer is $\boxed{\frac{3}{16}}$.
The line integral $\int_c \mathbf{F} \cdot d\mathbf{l}$ is the work done by the force $\mathbf{F}$ in moving a particle along the curve $c$. In this case, the force $\mathbf{F}$ is given by
$$\mathbf{F} = (x^2 – 2y) \hat{\imath} – 4yz \hat{\jmath} + 4xz^2 \hat{k}$$
and the curve $c$ is the straight line from $(0, 0, 0)$ to $(1, 1, 1)$.
To evaluate the line integral, we can use the following formula:
$$\int_c \mathbf{F} \cdot d\mathbf{l} = \int_a^b F_x dx + F_y dy + F_z dz$$
where $a$ and $b$ are the endpoints of the curve $c$. In this case, $a = (0, 0, 0)$ and $b = (1, 1, 1)$. Substituting in the expression for $\mathbf{F}$ and evaluating the integral, we get
$$\begin{align}
\int_c \mathbf{F} \cdot d\mathbf{l} &= \int_0^1 (x^2 – 2y) dx – 4yz dy + 4xz^2 dz \
&= \left[ \frac{x^3}{3} – 2xy \right]_0^1 – 4y \left[ \frac{z^2}{2} \right]_0^1 + 4x \left[ \frac{z^3}{3} \right]_0^1 \
&= \frac{1}{3} – 2(0) – 4(0) + \frac{4}{3} \
&= \frac{3}{16}
\end{align}$$
Therefore, the value of the line integral is $\boxed{\frac{3}{16}}$.
The other options are incorrect because they do not give the correct value of the line integral.