Consider the three alkyl halides labelled as I, II, and III in the...

Structure I is bromoethane, which is a primary ($1^\circ$) alkyl halide.

Structure II is 2-bromopropane, which is a secondary ($2^\circ$) alkyl halide.

Structure III is 2-bromo-2-methylpropane, which is a tertiary ($3^\circ$) alkyl halide.

The rate of an $\mathrm{S}_{N}2$ reaction depends on the steric hindrance around the electrophilic carbon atom. As the number of bulky alkyl groups attached to the reacting carbon increases, the approach of the nucleophile is hindered, leading to a decrease in reactivity.

Therefore, the general order of reactivity of alkyl halides towards $\mathrm{S}_{N}2$ reactions is $1^\circ > 2^\circ > 3^\circ$.

Following this order, the decreasing order of reactivity for the given compounds is $\text{I} > \text{II} > \text{III}$.

Answer: $\text{I} > \text{II} > \text{III}$