= Solution
If “steady <temperature> in 2050” means preventing the next annual increase relative to the climate at the delayed start, the relevant forcing increment is
$$
\Delta F_{2050}=5.4\ln(502.5/500)=0.0269327\ \mathrm{W\,m^{-2}},
\qquad
\boxed{m_{2050,\rm increment}=1.966\times10^3\ \mathrm{tons\ of\ SO_2}.}
$$
Its smaller value follows from $dF/dC=5.4/C$: the same additional ppm gives less incremental forcing at a higher baseline.
That interpretation must be separated from the earlier objective of holding <temperature> at the original reference climate. To cancel the accumulated increase from $415$ to $500$ ppm requires
$$
\Delta F_{\rm accumulated}=5.4\ln(500/415)=1.00618\ \mathrm{W\,m^{-2}},
\qquad
\boxed{m_{2050,\rm original\ reference}=7.344\times10^4\ \mathrm{tons\ of\ SO_2}.}
$$
This <annual versus accumulated radiative forcing> distinction is essential: the first box offsets just the further year's growth; the second is the equivalent one-year <aerosol> inventory needed for the accumulated forcing. In subsequent years that inventory must be replaced as particles disappear and increased as <carbon dioxide> rises. A delayed start does not erase the earlier warming merely because the annual incremental requirement is smaller.
\Image[/past-exam-of-the-chemistry-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-id-forcing-reference.png]
{title=Original comparison of annual and accumulated forcing and their distinct SO2 compensation masses}
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