 Solution

With $\gamma =2$ and $T(t+s)=T(t)+\dot{T}s$, the marginal damage is $2\kappa D_{0}W(t)e^{gs}[T+\dot{T}s]$. Integrating the constant and linear terms givesThe prefactor is $2(0.5\times 10^{-12})(0.003)(90\times 10^{12})=0.27$ in the corresponding dollar–temperature–time units. For the historical scenario with $r=0.03$ and $g=0.02$, the bracket is $1/0.01+0.02/0.01^2=300$, so $\boxed{S=81\mathrm{USD}\mathrm{per}\mathrm{tonne}\mathrm{C}{\mathrm{O}}_2}$. For $r=0.07$, $r-g=0.05$, giving global $S=0.27(20+8)=7.56$ dollars per tonne. Restricting impacts to ten per cent gives $\boxed{S_{\mathrm{domestic}}=0.756\mathrm{USD}\mathrm{per}\mathrm{tonne}\mathrm{C}{\mathrm{O}}_2}$.

The factor of about 107 reduction comes from both the higher discount rate and the geographical restriction; it is not a change in climate physics. The administration descriptions are those of the 2019 question, not claims about present policy. With a net discount rate of only one per cent, a century-scale horizon contributes substantially, making the linear warming and exponential income extrapolations consequential assumptions.