Step 1: Reverse the elimination:
Ethene plus $\text{C}_2\text{H}_5\text{N(CH}_3)_2$ recombine to a cation with two ethyl and two methyl groups on N.
Step 2: Consider the reagent:
Moist $\text{Ag}_2\text{O}$ is added to a halide salt, precipitating silver halide and leaving the quaternary hydroxide in solution. So the starting material must be a halide.
Step 3: Conclusion:
The halide of that cation is diethyldimethylammonium halide, option (A). The ethyltrimethyl cations would give trimethylamine instead.
Final Answer:
Substrate A is diethyldimethylammonium halide, option (A).
\[ \boxed{\text{Diethyldimethylammonium halide (A)}} \]