Chomsky Normal Form (If all of its production rules are of the form):
$A \rightarrow BC$ or
$A \rightarrow a$ or
$S \rightarrow \varepsilon$
where $A, B$ and $C$ are nonterminal symbols, $a$ is a terminal symbol ($a$ symbol that represents a constant value), $S$ is the start symbol, and $\varepsilon$ is the empty string. Also, neither $B$ nor $C$ may be the start symbol, and the third production rule can only appear if $\varepsilon$ is in $L(G)$, namely, the language produced by the context-free grammar $G$.
Applying productions of the first form will increase the number of nonterminals from $k$ to $k + 1$, since you replace one nonterminal $(-1)$ with two nonterminals $(+2)$ for a net gain of $+1$ nonterminal. Since you start with one nonterminal, this means you need to do $l - 1$ productions of the first form. You then need $l$ more of the second form to convert the nonterminals to terminals, giving a total of $l + (l - 1) = 2l - 1$ productions.
Correct Answer: $C$