Given that the collector current $I_C = 24\text{ mA}$, and it consists of $80\%$ of the electrons released by the emitter. This means $I_C = 0.8 I_E$. We can calculate the emitter current: $I_E = \frac{I_C}{0.8} = \frac{24\text{ mA}}{0.8} = 30\text{ mA}$. The base current is the difference between the emitter and collector currents: $I_B = I_E - I_C = 30\text{ mA} - 24\text{ mA} = 6\text{ mA}$. In an n-p-n transistor, the conventional current direction is opposite to the flow of electrons. Since electrons flow from the emitter (n-type) to the base (p-type) and then to the collector, the conventional current flows into the collector and base, and out of the emitter. Therefore, the base current is entering the base.