When 0.560 g of na(s) reacts with excess f2(g) to form naf(s), 13.8 kj of heat is evolved at standard-state conditions. what is the standard enthalpy of formation ( ) of naf(s)?

Enthalpy of formation is the alteration in enthalpy for the formation of one mole of the compound; hence you just have to modify the data you have proportionately to what it would be for one mole of the compound. 
Begin with the balanced equation first, just to safeguard the coefficient on the substance you're given is 1, remembering that the coefficient on your product must be 1. 
Na + (1/2)F2 --> NaF So no problem, since the coefficient of Na is 1. Change the mass of Na you're given to moles: (0.560 g Na)(1 mol Na/22.99 g Na) = 0.0244 mol Na. So, 13.8 kJ heat is altered by the response of 0.0244 mol Na.To look for the heat evolved by 1 mol Na, just divide the amount of heat by the number of moles: 13.8 kJ/0.0244 mol which gives 566.5 kJ/mol. Lastly, whether the heat is changed or obligatory to control the sign of ΔH. Heat is evolved here, meaning the reaction is exothermic, so ΔH = -566 kJ/mol.

Most likely a is the answer - because  you generate   38 atp's but also consume an atp.   each atp gives approx. less than 4 kcal so probably a is the closest answer.