Straight from Ebers-Moll:
$$ \frac{I_{C_2}}{I_{C_1}}=\exp\left(\frac{\Delta V_{_\text{BE}}}{V_T}\right)$$
(You should be able to derive this.)
So, assuming about \$-2.1\:\text{mV}\$ change in \$V_{_\text{BE}}\$ per Kelvin rise (see citation below), one would expect to see a \$-16.8\:\text{mV}\$ change for an \$8^\circ\text{C}\$ rise. I believe that AofE also uses \$V_T=25\:\text{mV}\$ (see citation below), so this says I should expect to see \$ \frac{I_{C_2}}{I_{C_1}}\approx 1.96\$.
As the starting assumption says \$I_{_\text{C}}=1\:\text{mA}\$, you should then find the new \$I_{_\text{C}}\approx 1.96\:\text{mA}\$. That will produce a drop across the \$10\:\text{k}\Omega\$ collector resistor of \$19.6\:\text{V}\$. That will cause \$V_{_\text{CE}}=400\:\text{mV}\$. And that is in saturation.
This is the direct use of Ebers-Moll by the authors used to make their statement.
supporting notes from AofE, 3rd edition
and,
