To find the permittivity of the surface when a wave is incident at an angle of 60 degrees and reflected at an angle of 45 degrees in air, we can use the laws of reflection and Snell's law.

1. Laws of Reflection:
According to the laws of reflection, the angle of incidence (θi) is equal to the angle of reflection (θr). In this case, the angle of incidence is 60 degrees and the angle of reflection is 45 degrees.

2. Snell's Law:
Snell's law relates the angles and velocities of a wave as it passes from one medium to another. It can be expressed as:
n1 * sin(θi) = n2 * sin(θr)
where n1 and n2 are the refractive indices of the two media, and θi and θr are the angles of incidence and refraction, respectively.

3. Refractive Index:
The refractive index of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in the medium. Since the wave is incident and reflected in air, the refractive index of air is approximately 1.

4. Applying Snell's Law:
Using Snell's law, we can rewrite the equation as:
sin(60) = n2 * sin(45)
sin(60) = n2 * (sqrt(2)/2)
n2 = sin(60) / (sqrt(2)/2)
n2 = (sqrt(3)/2) / (sqrt(2)/2)
n2 = sqrt(3/2)

5. Permittivity:
The permittivity (ε) of a medium is related to the refractive index by the equation:
n = sqrt(εr)
where εr is the relative permittivity of the medium.

6. Solving for Permittivity:
In this case, we are interested in the permittivity of the surface, which can be denoted as εs. Since the wave is incident and reflected in air (which has a relative permittivity of 1), we can write:
n2 = sqrt(εs)
(sqrt(3/2))^2 = εs
3/2 = εs
εs = 1.5

Therefore, the permittivity of the surface is 1.5, which corresponds to option 'D'.