What is the standard enthalpy of formation of methane, given that the average $ C - H $ bond enthalpy is $ 41KJmo{l^{ - 1}} $ and the reactions: $ {C_{(s)}} \to {C_{(g)}} $ , $ \Delta {H^ \circ } = 716KJmo{l^{ - 1}} $ , $ 2{H_{2(g)}} \to 4{H_{(g)}},\Delta {H^ \circ } = 872.8KJmo{l^{ - 1}} $ .

Hint: Enthalpy of formation is the enthalpy change for the formation of $ 1mol $ of a compound from its component elements, such as the formation of carbon dioxide from carbon and oxygen. The standard enthalpy of formation is measured in units of energy per amount of substance, usually stated in kilojoule per mole $ (KJmo{l^{ - 1}}) $ .

Complete answer:
 The standard enthalpy of formation or standard heat of formation $ (\Delta {H^ \circ }_f) $ of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.
Here, for the reaction, the standard enthalpy of formation for methane is:
 $ C + 2{H_2} \to C{H_4} $
Here, we have to find the change in enthalpy for the given reaction by using Hess law. So, the hess’s law states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function.
It is given that the average $ C - H $ bond enthalpy is $ 414KJmo{l^{ - 1}} $ . This is an endothermic process and the given value is positive.
However, we are forming methane. So, the change in enthalpy for the reaction will be positive.
Here, we are making four $ C - H $ bond
So, $ \Delta {H^ \circ }_{rxn1} = 4 \times ( - 414KJmo{l^{ - 1}}) $ .
 $ = - 1656KJmo{l^{ - 1}} $
Now, for the other two reactions that are given
 $ {C_{(s)}} \to {C_{(g)}} $
$ = \Delta {H^ \circ }_{rxn2} = 716KJmo{l^{ - 1}} $
 And,
 $ 2{H_{2(g)}} \to 4{H_{(g)}} $
 $ = \Delta {H^ \circ }_{rxn3} \to 872.8KJmo{l^{ - 1}} $
Since Hess' Law tells you that the overall enthalpy change for a reaction is independent of the pathway or the number of steps taken.
Now we add all $ 3 $ equations, we get.
 $ \Delta {H^ \circ }_f = \Delta {H^ \circ }_{rxn1} + \Delta {H^ \circ }_{rxn2} + \Delta {H^ \circ }_{rxn3} $ .
 $ \Delta {H^ \circ }_f = 716KJmo{l^{ - 1}} + ( - 1656KJmo{l^{ - 1}}) + 872.2KJmo{l^{ - 1}} $
 $ \Delta {H^ \circ }_f = - 67.2KJmo{l^{ - 1}} $ .

Note:
Hess's law is due to enthalpy being a state function, which allows us to calculate the overall change in enthalpy by simply summing up the changes for each step of the way, until the product is formed. The principle underlying Hess's law does not just apply to Enthalpy and can be used to calculate other state functions like changes in Gibbs' Energy and Entropy.