The correct answer is A. 36.2psnm-1 km-1.
The total first order dispersion is the sum of the material, waveguide, and profile dispersions. The material dispersion is the change in the refractive index with respect to the wavelength of light. The waveguide dispersion is the change in the refractive index with respect to the frequency of light. The profile dispersion is the change in the refractive index with respect to the position of the light in the waveguide.
The material dispersion is given by the following equation:
$$\frac{d\beta}{d\lambda} = -\frac{n_2}{c}$$
where $n_2$ is the group refractive index and $c$ is the speed of light in a vacuum.
The waveguide dispersion is given by the following equation:
$$\frac{d\beta}{d\omega} = -\frac{n_g}{c}$$
where $n_g$ is the group refractive index of the waveguide.
The profile dispersion is given by the following equation:
$$\frac{d\beta}{dz} = -\frac{n_p}{c}$$
where $n_p$ is the profile refractive index.
The total first order dispersion is given by the following equation:
$$\frac{d\beta}{d\lambda} = \frac{d\beta}{d\omega} + \frac{d\beta}{dz}$$
Substituting the equations for the material, waveguide, and profile dispersions into this equation, we get the following equation:
$$\frac{d\beta}{d\lambda} = -\frac{n_2}{c} – \frac{n_g}{c} – \frac{n_p}{c}$$
The values for the material, waveguide, and profile dispersions are given in the question. Substituting these values into the equation, we get the following equation:
$$\frac{d\beta}{d\lambda} = -\frac{2.8}{c} + \frac{20.1}{c} + \frac{23.2}{c} = 36.2\text{ ps/nm/km}$$
Therefore, the total first order dispersion is 36.2psnm-1 km-1.