Question
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The element whose II (I.P.) is less than its I (I.P.)
A.Li
B.Be
C.He
D.None

Answer
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Hint:
Mostly the second ionization potential tends to be higher than the first because an electron requires an extremely large amount to be removed from unipositive cation due the presence of strong electrostatic force of attraction between them. Unipositive cations are formed by loss of an electron and has charge of 1+

Complete step by step answer:
The electronic configuration of the elements mentioned in the question, i.e. Helium, Lithium and Beryllium are as follows:
He - $1{s^2}$
Li - $1{s^2}2{s^1}$
Be - $1{s^2}2{s^2}$
Over here we can see that the valence shell or the last shell is either half full or completely full. This means that all these elements are very stable. And as discussed earlier, ionization potential (IP) is remarkably high if the element is stable. Hence, the value of Ionization Potentials is going to be extremely high for these elements as compared to other lesser stable elements.
But the real trick with these elements arises after the first ionization takes place. When the first ionization takes place, the electronic configurations of the given elements will be as follows:
He - $1{s^1}$
Li - $1{s^2}2{s^0}$
Be - $1{s^2}2{s^1}$
This shows that all the elements become unipositive cations. And if we observe the atomic numbers of these elements, we can conclude that the ratio of the negative charge to the positive charge in the atom reduces drastically. This results in an even stronger pull due to the positive charge from the nucleus. This indirectly means that when we would try ionizing these elements for the second time, it would be even more difficult for 2 reasons:
Even after these elements are ionized for the first time, the resulting ionized electronic configurations are again with half-filled or completely filled valence shells, which makes them stable again.
There is a drastic decrease in the ratio of negative charge to the positive charge from the nucleus, resulting in a greater pull/attraction from the nucleus.
Due to these reasons, we can conclude that neither of the mentioned elements He, Li and Be, will have a II (I.P.) which is less than its I (I.P.)

Hence, Option D is the correct option.

Note:
Ionization potential is also known as ionization energy. The unit of I.P. is eV/atom or KJ/gm-atom. I.P. decreases as we move down the group due to the increase in number of shells. When comparing the Ionization Potentials, He has the highest IP of all elements.