Answer
The total mass of the plutonium in the warhead was $1.3 \times 10^{3} \mathrm{kg}$
Work Step by Step
Each warhead would have yielded 2.0 Megaton of TNT.
1 TNT = 2.6 x $10^{28}$ MeV energy
2 Megaton of TNT releases energy of
2 * 2.6 x $10^{28}$ MeV = 5.2 x $10^{28}$ MeV
One fission event releases 200MeV of energy. To count how many fission events happened, we can simply perform the calculation as below,
$N=\frac{5.2 \times 10^{23} \mathrm{MeV}}{200 \mathrm{MeV}}=2.6 \times 10^{26}$
But, this number only denotes only 8% of the entire Pu present in the warhead that undergoes fission and therefore, the amount of Plutonium present in the warhead could be given by;
$N_{0}=\frac{N}{0.080}=\frac{2.6 \times 10^{26}}{0.080}=3.25 \times 10^{27}=5.4 \times 10^{3} \mathrm{mol}$
For Plutonium, the weight per mole is;
M = 0.239 kg/mol
As we know the Molecular weight and the number of moles of Plutonium in the warhead, we can calculate the mass of the warhead by simple calculation;
$m=\left(5.4 \times 10^{3} \mathrm{mol}\right)(0.239 \mathrm{kg} / \mathrm{mol})=1.3 \times 10^{3} \mathrm{kg}$
