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Suggest and explain an indirect method to calculate lattice enthalpy of magnesium bromide.

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Born Haber.s cycle method:
`Mg(s)+Br_(2)(l) to MgBr_(2)(s) " " DeltaH_f^0`
Sublimation : `Mg_((s)) to Mg_((g)) " " DeltaH_1^0`
Ionisation : `Mg_((g)) to Mg_((g))^(2+) + 2e^(-) " " DeltaH_2^0`
Vapourisation : `Br_(2(l)) to Br_(2(g)) " " DeltaH_3^0`
Dissociation : `Br_(2(g)) to 2Br_((g)) " " DeltaH_4^0`
Electron affinity : `2Br_((g)) + 2e^(-) to 2Br_((g))^(-) " " DeltaH_5^0`
Lattice enthalpy : `Mg_((g))^(2+) + 2Br_((g))^(-) to MgBr_((2)s) " " DeltaH_6^0` =?
`DeltaH_f^0 = DeltaH_1^0+DeltaH_2^0 + DeltaH_3^0+ DeltaH_4^0 + DeltaH_5^0 + DeltaH_6^0`
`DeltaH_6^0=DeltaH_f^0 -(DeltaH_1^0+DeltaH_2^0+DeltaH_3^0+DeltaH_4^0+DeltaH_5^0)`
If we know the values of `DeltaH_f^0,DeltaH_1^1,DeltaH_2^0, DeltaH_3^0, DeltaH_4^0` and `DeltaH_5^0` . We can calculate the value of `DeltaH_6^0` by indirect method.
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When a mole of magnesium bromide is prepared from 1 mole of magnesium and 1 mole of liquid bromine, 524 kJ of energy is released. The heat of sublimation of Mg metal is 148 "kJ mol"^(-1) . The heat of dissociation of bromine gas into atoms is 193 "kJ mol"^(-1) . The heat of vaporization of liquid bromine is 31 kJ "mol"^(-1) . The ionisation energy of magnesium is 2187 kJ mol^(-1) and the electron affinity of bromine is - 662 "kJ mol"^(-1) . Calculate the lattice energy of magnesium bromide.