Step 1: Count N-H bonds
Ethylamine has two N-H bonds, so it takes two methyl groups to make a tertiary amine.
Step 2: One more step
The lone pair of the tertiary amine attacks one more CH$_3$I, giving the quaternary salt $[\text{C}_2\text{H}_5\text{N(CH}_3)_3]^+\text{I}^-$, option (B).
Final Answer:
The quaternary salt ethyltrimethylammonium iodide forms, option (B).
\[ \boxed{\text{Ethyltrimethylammonium iodide}} \]