 Solution

Albedo reduction increases absorbed short-wave flux by $S_0(a_{1}T+a_2{T}^2)/4$. Define $A={\lambda }_P-1.5-S_0{a}_1/4$ and $B=S_0{a}_2/4$. The perturbation energy balance is $N(T)=F-AT+B{T}^2$, where $F=3.7$ W m${}^{-2}$ for doubling. With the preceding constants, $A\simeq 1.220$ W m${}^{-2}$ K${}^{-1}$ and $B=0.068$ W m${}^{-2}$ K${}^{-2}$. The two roots areStability follows from $N^{\prime }(T)=-A+2BT<0$ at the lower root. They merge at $T=A/(2B)\simeq 8.97$ K and $F_{\max }=A^2/(4B)\simeq 5.47$ W m${}^{-2}$; at equality the equilibrium is marginal, not linearly stable.

Assuming the usual logarithmic forcing extension $F=3.7\log (C/280)/\log 2$ gives a stable branch only for $\boxed{C<C_{\mathrm{crit}}\simeq 7.8\times 10^2\mathrm{ppm}}$. This “maximum” is the supremum of stable concentrations in the stated quadratic, linearized model. It is not a physical prediction of a real-world runaway threshold; extrapolating its feedback laws far beyond their small-change regime is an additional assumption.