Relative reactivities of alkyl halides in an E2 reaction
I believe you might be mixing up E1 and E2 elimination reactions. In the first case the reaction first proceeds by formation of a carbocation from the alkyl halide and then a proton extractio. In this case, the order of reactivity is indeed $3^∘&gt;2^∘&gt;1^∘$ as such is the order of carbocation stability - in order the E1 reaction to proceed the carbocation must be formed.

But the situation is completely different in an E2 elimination. In this case the mechanism is the abstraction of the proton that is oriented anti to the halide with the simultaneous formation of the double bond and halide leaving the molecule as anion. That means that in this case the stability of carbocation plays no role and the significant factor are the sterical considerations. If the carbon atoms around the halide make it hard for the base to get to the anti proton then the reaction is less likely to proceed via E2 mechanism.

But bear in mind that usually both E1 and E2 elimination can take place and because of that tertiary halides would usually undergo elimination more readily than primary halides - just not via E2 mechanism.

More details on the E2 mechanism, with pictures: Master Organic Chemistry