let T(n) is the time taken by nth full adder as it has to wait for 14ns(PD$_C$) for previous adder to calculate its carry
so we can form a recurrence
T(n) = T(n-1) + PD$_C$ (Propagation delay of carry)
T(1) = (PD$_s$) Propagation delay of sum (as for only 1 full adder time taken = 44ns(=(PD$_s$)
solving by back substitution
T(n) = T(n-2) + (PD$_C$)+ (PD$_C$)
T(n) = T(n-2) + 2*(PD$_C$)
T(n) = T(n-k) + k*(PD$_C$)
if n-k=1 ,k=n-1
T(n) = T(n-(n-1)) + (n-1)*(PD$_C$)
T(n) = T(1) + (n-1)*(PD$_C$)
T(n) = (PD$_C$) + (n-1)*(PD$_C$) we can derive this formula by ton a ways
here n=4 (PD$_C$) = 14ns (PD$_s$) =44ns
T(4) = 44 + (4-1)*(14)
T(4) =86ns