In NPN transistor the collector current is 8mA.  If 80% of electrons emitted reach the collector, then

$\begin{array}{l}Collector Current :\hspace{0.17em}{I}_c=8mA\\ \\ Given :\\ 80\% of emitted electrons = Collected by collector\\ \Rightarrow 0.8\times I_E = I_C\\ \Rightarrow I_E=\frac{I_c}{0.8}=\frac{8mA}{0.8}=10mA\\ Now, \\ I_E =I_B + I_c\\ \Rightarrow I_B=I_E-I_c=10-8=2mA\end{array}$