A common emitter amplifier circuit is shown in the figure below. For the transistor used in the circuit the current amplification factor, βdc=100. Other parameters are mentioned in the figure.We find that :

βdc  =  100,               βdc=iCiB  ⇒   iB= ic100   −−−−(1)VCC =  +iCRL  + VBC  +  VBE       −−−−−−(2)−iBRB  +  VBC  +  iCRL=0        −−−−−−(3)Solving we get   VBE=+20.7V,     VBC=−3.75VNOTE: According to the question none of the options are matching. 4 th option is nearest.