Answer
1100 kg.
Work Step by Step
As stated in several previous problems, assume that 200 MeV is released in one fission of a uranium nucleus.
We are told to assume an efficiency of $34\%$. This means that fission actually generates a power of 950MW/0.34.
Now that we know the average power for a year, find the amount of energy.
$$\frac{(950\times10^6 J)/0.34}{s}\frac{3.156\times10^7s}{year}\frac{MeV}{1.60\times10^{-13}J}=5.511\times10^{29}MeV/y$$
Change this to an amount of uranium.
$$\frac{ 5.511\times10^{29}MeV }{y}\frac{1fission}{200\;MeV}\frac{0.235kg}{6.02\times10^{23}atoms}$$
$$=1076kg\approx 1100kg$$
