To determine the air-fuel ratio by weight for the combustion of methane, we need to understand the stoichiometry of the reaction.

1. Stoichiometry of Methane Combustion:
Methane (CH4) reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The balanced chemical equation for this reaction is:

CH4 + 2O2 -> CO2 + 2H2O

From the equation, we can see that one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water.

2. Atomic Mass Calculation:
To determine the air-fuel ratio by weight, we need to calculate the atomic masses of carbon (C), hydrogen (H), and oxygen (O).

- Atomic mass of Carbon (C) = 12.01 g/mol
- Atomic mass of Hydrogen (H) = 1.008 g/mol
- Atomic mass of Oxygen (O) = 16 g/mol

3. Calculation of Molecular Weights:
Next, we calculate the molecular weights of methane (CH4) and oxygen (O2).

- Molecular weight of CH4 = (12.01 + 4 * 1.008) g/mol = 16.04 g/mol
- Molecular weight of O2 = 2 * 16 g/mol = 32 g/mol

4. Air-Fuel Ratio Calculation:
The air-fuel ratio by weight is the ratio of the mass of air to the mass of fuel required for stoichiometric combustion.

From the balanced chemical equation, we know that one mole of methane requires two moles of oxygen for complete combustion. Therefore, the stoichiometric ratio of methane to oxygen is 1:2.

Using the molecular weights calculated earlier, we can determine the mass ratio of methane to oxygen:

- Mass ratio of CH4 to O2 = (16.04 g/mol) / (32 g/mol) = 0.50125

Since the stoichiometric ratio of methane to oxygen is 1:2, the mass ratio of air to methane is twice the mass ratio of oxygen to methane:

- Mass ratio of air to CH4 = 2 * 0.50125 = 1.0025

Finally, the air-fuel ratio by weight is the inverse of the mass ratio of air to fuel:

- Air-fuel ratio by weight = 1 / 1.0025 ≈ 0.9975

Therefore, the correct answer is option 'D': 17.16.