So for branch instructions CPI will be 3 , as no of stalls are 2 , and there are 2*10^8 branch instructions , therefore total machine cycles used =2*10^8*3 and for remaining 8*10^8 instructions CPI will be 1 as they are non branch instructions , and total machine cycles taken =8*10^8*1,now total machine cycles used = 8*10^8*1+3*2*10^8 = 14*10^8 , now cpu frequency is 1Ghz Therefore memory cycle time =1/(10^9), finally total execution time = total no of machine cycles used *cycle time=(14*10^8)/10^9= 1.4 sec