A 1000 MW fission reactor consumes half of its fuel in 5 yr. How much ${}_{92}^{235}U$ did it contain initially?

A 1000 MW fission reactor consumes half of its fuel in 5 yr. How much ${}{92}^{235}U$ did it contain initially?
Assume that the reactor operates 80% of the time that all the energy generated - arises from the fission of ${}
{92}^{235}U$ and that this nuclide is consumed only by the fission process.

Given power of reactor P = 1000MW
We will use concept that the energy generated in one fission of ${}{92}^{235}U$ is 200 MeV.
Number of ${}
{92}^{235}U$ attom in 1 g = 1/235 x 6.023 x ${{10}^{23}}$
Energy generated per gram of ${}{92}^{235}U$ =(1/235 x 6.023x${{10}^{23}}$ x 200 x 1.6x${{10}^{-13}}$
Total energy generated in 5 year with 80% of the time = 1000 x ${{10}^{6}}$x5x365x24x60x60x80/100
Mass of ${}
{92}^{235}U$ consumed in 5 yr,

m = Total energy/Energy consumed per gram
= 1.538 x ${{10}^{6}}$ g = 1538 kg

Initial amount of ${}_{92}^{235}U$ = 1544 x 2 kg = 3076 kg