The correct answer is $\boxed{\text{B) 9.6ns km-1}}$.
The rms pulse broadening due to material dispersion is given by:
$$\Delta \tau = \frac{L}{\nu_0} \cdot \frac{D_m}{\lambda_0^2}$$
where $L$ is the fiber length, $\nu_0$ is the central frequency of the light, $D_m$ is the material dispersion coefficient, and $\lambda_0$ is the central wavelength of the light.
In this case, we are given that $L = 1 \text{ km}$, $\nu_0 = 299,792,458 \text{ m s}^{-1}$, $D_m = 150 \text{ ps nm}^{-1} \text{ km}^{-1}$, and $\lambda_0 = 600 \text{ nm}$. Substituting these values into the equation, we get:
$$\Delta \tau = \frac{1 \text{ km}} {299,792,458 \text{ m s}^{-1}} \cdot \frac{150 \text{ ps nm}^{-1} \text{ km}^{-1}} {600 \text{ nm}^2} = 9.6 \text{ ns km}^{-1}$$
Therefore, the rms pulse broadening due to material dispersion is $\boxed{\text{9.6ns km-1}}$.
The other options are incorrect because they do not take into account the correct values of the parameters.