First, second and third ionization enthalpies of Al_(g) are 577,1816 and 2744 kJ/mole. The amount of energy required to convert 6.02 x 10²²  gaseous Al atoms into gaseous Al⁺³ ion is 

$\mathrm{Al}\to {\mathrm{Al}}^{3+}+3{\mathrm{e}}^{-}$

Energy required to remove three electrons from Al is equal to addition of first, second and third ionization energies of Al. 

energy required = 577 (first ionization energy) + 1816 (second ionization energy) + 2744 (third ionization energy) = 5137 kJ

Energy required for one Al atom is $\frac{5137}{6.02\times {10}^{23}}$

Energy required for given atoms = $\frac{6.02\times {10}^{22}\times 5137}{6.02\times {10}^{23}}$=513.7 kJ