A change of 8.0 mA in the emitter current brings a change of 7.9 mA...

We have, $$\begin{aligned} \quad I_E & =I_B+I_C \\ \Delta I_E & =\Delta I_B+\Delta I_C\end{aligned}$$ According to question, $\text{ Thus, } \quad$$\begin{aligned} \Delta I_E & =8.0 \mathrm{~mA} \\ \Delta I_C & =7.9 \mathrm{~mA} \\ \Delta I_B & =8.0 \mathrm{~mA}-7.9 \mathrm{~mA} \\ & =0.1 \mathrm{~mA} \end{aligned}$$$Thus,$\quad \Delta I_B=8.0 \mathrm{~mA}-7.9 \mathrm{~mA}$Now,$\alpha=\frac{I_C}{I_E}=\frac{\Delta I_C}{\Delta I_E}=\frac{7.9 \mathrm{~mA}}{8.0 \mathrm{~mA}} \cong 0.99$