= Solution
Use the thermal power when applying <nuclear fuel burnup>. No <capacity factor> is supplied, so take a full-power year of $365$ days; if the operating fraction were $u$, the annual quantities in this part would be multiplied by $u$.
$$
P_{\rm th}=\frac{1650}{0.37}=4459.459\ \mathrm{MW},\qquad
E_{\rm th}=4459.459(365)=1.627703\times10^6\ \mathrm{MW\,d\,yr^{-1}}.
$$
The corresponding initial heavy-metal <nuclear fuel> requirement is
$$
M=\frac{E_{\rm th}}{55000}=\boxed{29.5946\ \mathrm{t\,U\,yr^{-1}}}.
$$
All <nuclear fuel> and concentrate masses below are <uranium> masses, excluding oxygen. Let $F,P,W$ denote <uranium> feed, product and <enrichment tails>, and $x_f,x_p,x_w$ the <uranium-235> <mass fractions>. Conservation of total <uranium> and <uranium-235> gives
$$
F=P+W,\qquad Fx_f=Px_p+Wx_w,
$$
so
$$
\frac FP=\frac{x_p-x_w}{x_f-x_w}
=\frac{0.050-0.003}{0.007-0.003}=11.75.
$$
With $P=M$, the <uranium oxide concentrate> requirement is therefore
$$
F=11.75M=\boxed{347.7365\ \mathrm{t\,U\,yr^{-1}}},\qquad
W=318.1419\ \mathrm{t\,U\,yr^{-1}}.
$$
This is not the gross mass of oxide concentrate. Both $F-P-W=0$ and $0.007F-0.050P-0.003W=0$ provide independent balance checks.
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