You are correct! You first calculated the kJ per 1 gram and then converted this answer to kJ per mole by using the following calculation...

1.23 kJ/gK x 1g x 6.12 K = -7.53kJ for the one gram Since there are 80 grams per mole of ammonium nitrate, the molar heat of decomposition will be 80 times this number which is -602 kJ/mole.