$Li, Be, B$ and $C$ are the elements of the same period of the Modern Periodic Table.
$(a)$ Arrange them in increasing order of their atomic size.
$(b)$ In which shell (number) would the last electron enter for all of them?
$(c)$ Calculate the number of valence electrons in each.
$(d)$ Which element amongst them is most electropositive?

(A) The atomic size decreases across a period from left to right. Therefore,the increasing order of atomic size is: $C < B < Be < Li$.
$(b)$ These elements belong to the $2nd$ period of the Modern Periodic Table. Therefore,the last electron enters the $L$ shell (the $2nd$ shell).
$(c)$ The number of valence electrons corresponds to the group number for these elements: $Li = 1, Be = 2, B = 3, C = 4$.
$(d)$ Electropositivity decreases across a period from left to right. Thus,$Li$ is the most electropositive element among them.