Chomsky normal form and regular languages
Assuming that $B$ is regular, then $L(G)=B \cup \lbrace w \in L(G) \mid \vert w \vert \leq 2^k \rbrace$ is also regular because the union of a finite number of regular languages is regular  (especially, $k$ is a constant as the number of variables in the Chomsky normal form). 

This can not be true in general because there are context free languages which are not regular.