Answer
$9.1\times10^{-4}\approx 0.09\%\approx\frac{1}{1100}$.
Work Step by Step
The fraction is 200 MeV divided by the mass-energy of a $^{235}_{92}U$ nucleus.
$$\frac{200\;MeV}{mc^2}$$
$$=\frac{200\;MeV}{(235\;u)\left(\frac{931.49MeV/c^2}{u}\right)}$$
$$=9.1\times10^{-4}\approx 0.09\%\approx\frac{1}{1100}$$
