Ans is 23ns.
Lets say every iPUSH and iPOP operation takes 'X' ns, where 'i' represents the element index.
Let 'Y' ns be the delay between two consecutive PUSH and POP operation.
Since it is a stack of 4elements. The operations will be in the order :
1PUSH -> 2PUSH -> 3PUSH -> 4PUSH -> 4POP -> 3POP -> 2POP -> 1POP
For 4th element, stack-life = Y
For 3rd element, stack-life = Y+4PUSH+Y+4POP+Y = 3Y +2X
For 2nd element, stack-life = Y+3PUSH+Y+4PUSH+Y+4POP+Y+3POP+Y= 5Y + 4X
For 1st element, stack-life = Y+2PUSH+Y+3PUSH+Y+4PUSH+Y+4POP+Y+3POP+Y+2POP+Y = 7Y + 6X
So, average stack-life of an element = ((Y)+(3Y+2X)+(5Y+4X)+(7Y+6X))/4 = (16Y + 12X)/4 = 4Y + 3X
Given in question, Y=2ns; X=5ns
Ans : ((4*2) + (3*5)) = 23ns