At STP, that is standard temperature and pressure conditions, 1 mol of any gas occupies a volume of 22.4 L
the balanced equation for the reaction between Li and N₂;
6Li + N₂ --> 2Li₃N
stoichiometry of Li to N₂ is 6:1
since 22.4 L is occupied by 1 mol
the number of moles occupying 0.0593 mL is - 1/22.4 x 0.0593 = 0.00265 mol
the number of N₂ moles reacted - 0.00265 mol
if 1 mol of N₂ reacts with 6 mol of Li
then 0.00265 mol of N₂ reacts with - 6 x 0.00265 = 0.0159 mol
the mass of Li reacted is 0.0159 mol x 7 g/mol = 0.111 g
mass of Li required - 0.111 g