Answer
$4\times10^{-4}$ grams, keeping one significant figure.
Work Step by Step
As stated in the 3 previous problems, assume that 200 MeV is released in one fission of a uranium nucleus.
Change the 30 million joules of energy to an amount of uranium.
$$(3\times10^7 J) \frac{MeV}{1.60\times10^{-13}J}\frac{1fission}{200\;MeV}\frac{235g}{6.02\times10^{23}atoms}$$
$$\approx 4\times10^{-4}g=4\times10^{-7}kg$$
