**Answer:**

To determine the air-fuel ratio by weight, we need to calculate the mass of air required to burn one unit mass of methane.

The balanced chemical equation for the combustion of methane (CH4) is as follows:

CH4 + 2O2 -> CO2 + 2H2O

From the balanced 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.

The molar mass of methane (CH4) is 16 g/mol, and the molar mass of oxygen (O2) is 32 g/mol.

Therefore, the mass of air required to burn one unit mass of methane can be calculated as follows:

Mass of air = (2 * Molar mass of oxygen) / Molar mass of methane
= (2 * 32 g/mol) / 16 g/mol
= 4 grams

So, the air-fuel ratio by weight is 4, which means that 4 grams of air is required to burn 1 gram of methane.

However, the given answer options do not include 4.

To convert the air-fuel ratio by weight to the air-fuel ratio by volume, we need to consider the molar volume of gases at standard temperature and pressure (STP).

At STP, one mole of gas occupies a volume of 22.4 liters.

The molar volume of air can be calculated as follows:

Molar volume of air = (2 * Molar volume of oxygen) + (79 * Molar volume of nitrogen)
= (2 * 22.4 liters/mol) + (79 * 22.4 liters/mol)
= 1592.8 liters/mol

The molar volume of methane can be calculated as follows:

Molar volume of methane = Molar volume of carbon dioxide
= 22.4 liters/mol

Therefore, the air-fuel ratio by volume can be calculated as follows:

Air-fuel ratio by volume = (Molar volume of air) / (Molar volume of methane)
= 1592.8 liters/mol / 22.4 liters/mol
= 71.2

The air-fuel ratio by volume is 71.2, which is equivalent to an air-fuel ratio by weight of 17.16 (since air is about 3.76 times heavier than methane).

Hence, the correct answer is option D) 17.16.