There are six different isomers of an alkyl bromide that can give 3-methyl-2-hexene as one of the elimination products by an E2 mechanism. Let's break down the explanation step by step:

1. Understand the reaction:
In an E2 elimination reaction, an alkyl halide reacts with a strong base (such as C2H5ONa) in the presence of an alcohol (C2H5OH) to form an alkene. The alkyl halide undergoes elimination of a halide ion and a proton from adjacent carbon atoms, resulting in the formation of a double bond.

2. Identify the alkyl bromide:
The alkyl bromide in question should have six carbon atoms, as the desired product is a 2-hexene with a methyl group at the third carbon. Let's consider the different possibilities for the alkyl bromide:

- n-Hexyl bromide: This is a straight-chain alkyl bromide with six carbon atoms. The elimination reaction would result in the formation of 3-methyl-2-hexene as one of the products.
- 2-Methylpentyl bromide: This is an alkyl bromide with a methyl group attached to the second carbon atom. The elimination reaction would also give 3-methyl-2-hexene.
- 3-Methylpentyl bromide: This is an alkyl bromide with a methyl group attached to the third carbon atom. The elimination reaction would give the desired product.
- 4-Methylpentyl bromide: This is an alkyl bromide with a methyl group attached to the fourth carbon atom. The elimination reaction would also give the desired product.
- 2,3-Dimethylbutyl bromide: This is an alkyl bromide with two methyl groups attached to the second and third carbon atoms. The elimination reaction would result in the formation of 3-methyl-2-hexene.
- 2,2-Dimethylbutyl bromide: This is an alkyl bromide with two methyl groups attached to the second and second carbon atoms. The elimination reaction would also give the desired product.

3. Count the isomers:
Considering all the possibilities mentioned above, we have identified six different isomers of alkyl bromide that can give 3-methyl-2-hexene as one of the elimination products.

Therefore, the correct answer is '6'.