This definitely seems incorrect.
If in the worst case we consider non-parallel execution of adders in ripple-ahead, even then it will require 4*20ns=80ns.
Whereas in parallel execution, it will be 50ns.
$c_{3}$ will reach last adder after 40ns and by that time we would have already done half of the sum in that last adder, so additional 10ns. Therefore 50ns.
In case of look-ahead generator we need 3 level and-or gate logic because of 2 fan-in.
Therefore, total 10+5+5+5+10 = 35ns.