The question "What is the volume of energy required to burn 45cm³ of mixture at S.T.P" contains a conceptual error, as energy is not measured in volume units (cm³). However, given the options are in c

The combustion of methane with oxygen produces carbon dioxide and water.

At Standard Temperature and Pressure (S.T.P.), the volumes of gaseous reactants and products are in simple whole-number ratios, which correspond to their stoichiometric coefficients in the balanced chemical equation.
From the balanced equation, the ratio of methane to oxygen is 1:2. This means 1 volume of $\text{CH}_4$ reacts with 2 volumes of $\text{O}_2$.

Given the volume of the "mixture" (assumed to be methane) is 45 cm³.
Volume of $\text{O}_2$ required = $2 \times \text{Volume of CH}_4$
Volume of $\text{O}_2$ required = $2 \times 45 \text{ cm}^3 = 90 \text{ cm}^3$

This result matches option D.

The final answer is $\text{D) 90.0cm}^3$.