Answer
84 kg
Work Step by Step
The bomb releases 66 kilotons times the amount of energy released in terms of mass of TNT.
One megaton of TNT releases $2.6 \times 10^{28} \mathrm{MeV}$ of energy.
$\therefore$ 1 kilo ton of TNT releases $2.6 \times 10^{28} \times 10^{-3} \mathrm{MeV} = 2.6 \times 10^{25}$ MeV of energy.
The energy yield of the bomb is given by
$$ E = \text{Amount of energy released in terms of TNT} \times \text{66}$$
$$
E=\left(66 \right)\left(2.6 \times 10^{25} \mathrm{MeV} / \text { megaton }\right)=1.72 \times 10^{27} \mathrm{MeV}
$$
About $200 \mathrm{MeV}$ is released in every fission event.
The total number of fission events taking place if we consider that 200 MeV is released in every fission event is given by
$$ \text{No of fission events} = \dfrac{\text{Energy released by bomb}}{\text{Energy relaeased per fission event}} $$
$$
=\left(1.72 \times 10^{27} \mathrm{MeV}\right) /(200 \mathrm{MeV})=8.58 \times 10^{24}
$$
We have used only 4% of 235U nuclei in fission,
The number of 235 U nuclei is given by
$$ \text{No of 235U nuclei} = \dfrac{\text{Number of fission events}}{\text{percentage of total Uranium nuclei used in fission}} $$
$\left(8.58 \times 10^{24}\right) /(0.040)=2.14 \times 10^{26}$
The mass of Uranium in the atomic bomb can be calculated by multiplying the atomic mass
$$
\left(2.14 \times 10^{26}\right)(235 \mathrm{u})\left(1.661 \times 10^{-27} \mathrm{kg} / \mathrm{u}\right)=83.7 \mathrm{kg} \approx 84 \mathrm{kg}
$$
