The correct answer is $\boxed{\text{B) }2}$.
The line integral is a way of calculating the work done by a force along a curve. In this case, the force is given by the vector field $\mathbf{F}(x, y, z) = (2xy^2, 2x^2y, 0)$, and the curve is the line segment from the origin to the point $(1, 1, 1)$.
To calculate the line integral, we first need to find a parametrization of the curve. A convenient parametrization is $x = t$, $y = t$, and $z = t$, for $0 \leq t \leq 1$. This parametrization tells us that the point $(x, y, z)$ is on the curve when $t = 0$, it is at the origin, and it is at the point $(1, 1, 1)$ when $t = 1$.
Now we can calculate the line integral:
$$\int\limits_{\text{c}} {\left( {2{\text{x}}{{\text{y}}^2}{\text{dx}} + 2{{\text{x}}^2}{\text{ydy}} + {\text{dz}}} \right)} = \int\limits_0^1 {\left( {2t^2} \right)} dt = 2.$$
Therefore, the value of the line integral is $\boxed{\text{B) }2}$.
The other options are incorrect because they do not give the correct value of the line integral.