 Solution

In strict Chomsky normal form, every production of a context-free grammar is $A\to BC$ or $A\to a$, with $A,B,C$ nonterminal symbols and $a$ a terminal symbol. Such a derivation cannot produce the empty word: it starts with one nonterminal symbol, and every step either increases the symbol count or replaces one symbol by one symbol. The context-free language $\{\epsilon \}$, generated by $S\to \epsilon$, is therefore not generated by any strict CNF grammar.

The convention matters: extended Chomsky normal form permits $S\to \epsilon$ with $S$ excluded from right-hand sides. Under that convention every context-free language has such a grammar, so the requested example would not exist.