(d) Given $i_{c} = \frac{80}{100} \times i_{e} \Rightarrow 24 = \frac{80}{100} \times i_{e} \Rightarrow i_{e} = 30\ mA$ By using $i_{e} = i_{b} + i_{c} \Rightarrow i_{b} = 30 - 24 = 6\ mA$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.