= 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> $\{\varepsilon\}$, generated by $S\to\varepsilon$, is therefore \b[not generated by any strict CNF grammar].
The convention matters: extended <Chomsky normal form> permits $S\to\varepsilon$ 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.
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