= Solution
Use the recovered <uranium> inventory, rather than the initial fuel mass, as the recycle feed. Its available mass is
$$
F_r=33(1-0.015)=32.505 \mathrm{t\,U\,yr^{-1}}.
$$
Assuming the same <enrichment tails> assay and the specified higher recycled-fuel assay, the <isotopic mass balance> gives
$$
P_r=F_r\frac{0.0085-0.003}{0.043-0.003}=4.4694375 \mathrm{t\,U\,yr^{-1}},\qquad W_r=28.0355625 \mathrm{t\,U\,yr^{-1}}.
$$
Treat this fuel as replacing the same heavy-metal mass of ordinary fresh fuel: the stipulated higher <uranium-235> content compensates for <uranium-236>. The displaced natural feed is
$$
\Delta F=9.25P_r=\boxed{41.3423 \mathrm{t\,U\,yr^{-1}}},
$$
about \b[12.63% of the once-through natural-uranium requirement]. The residual natural feed demand is $286.052 \mathrm{t\,U\,yr^{-1}}$.
For separative work accounting define
$$
V(x)=(1-2x)\ln\frac{1-x}{x},\qquad S=PV(x_p)+WV(x_w)-FV(x_f).
$$
Using kilograms in this expression gives kg-SWU. For ordinary fresh fuel the work per product kilogram is
$$
s_f=V(0.040)+8.25V(0.003)-9.25V(0.007)=5.34661 \mathrm{kg\!\text{-}\!SWU\,kg^{-1}}.
$$
The fresh work avoided by substituting $P_r$ is $23896.3 \mathrm{kg\!\text{-}\!SWU\,yr^{-1}}$, but producing that recycled product still requires
$$
S_r=1000\big[P_rV(0.043)+W_rV(0.003)-F_rV(0.0085)\big]
=22409.6 \mathrm{kg\!\text{-}\!SWU\,yr^{-1}}.
$$
Hence the \b[net] saving is
$$
\Delta S=1000P_rs_f-S_r=\boxed{1486.73 \mathrm{kg\!\text{-}\!SWU\,yr^{-1}}}.
$$
The work saving is only about \b[0.786%] of the original annual separative work. Crediting the whole avoided fresh-fuel work without charging the recycle work would substantially overstate the saving. This is an annual resource account at the supplied assays, with no isotope-separation equipment model.
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