To determine the number of isomeric alcohols of molecular formula C6H14O that give a positive iodoform test, we need to consider the structural requirements for iodoform formation.

Iodoform is a yellow solid that is produced when a methyl ketone or an alcohol with a methyl group attached to a carbonyl carbon reacts with iodine and sodium hydroxide (NaOH). The reaction proceeds through the oxidation of the methyl group to a carboxylate ion, followed by the substitution of iodine for the carboxylate.

The structural requirements for iodoform formation are as follows:
1. The compound must contain a methyl group (CH3) attached to a carbonyl carbon (C=O).
2. The compound must have a hydrogen atom attached to the carbon adjacent to the carbonyl carbon (α-carbon).

Now, let's analyze the possible isomeric alcohols of molecular formula C6H14O and determine which ones meet these requirements.

Possible isomeric alcohols of molecular formula C6H14O:
1. Hexanol (C6H14O): This alcohol does not have a methyl group attached to a carbonyl carbon, so it does not meet the first requirement for iodoform formation.

2. 2-Methylpentanol (C6H14O): This alcohol has a methyl group attached to a carbonyl carbon and a hydrogen atom attached to the α-carbon, so it meets both requirements for iodoform formation.

3. 3-Methylpentanol (C6H14O): This alcohol has a methyl group attached to a carbonyl carbon and a hydrogen atom attached to the α-carbon, so it meets both requirements for iodoform formation.

4. 4-Methylpentanol (C6H14O): This alcohol has a methyl group attached to a carbonyl carbon and a hydrogen atom attached to the α-carbon, so it meets both requirements for iodoform formation.

5. 2,2-Dimethylbutanol (C6H14O): This alcohol does not have a methyl group attached to a carbonyl carbon, so it does not meet the first requirement for iodoform formation.

Therefore, out of the given isomeric alcohols, only 2-Methylpentanol, 3-Methylpentanol, and 4-Methylpentanol meet the structural requirements for iodoform formation. Hence, the correct answer is option 'C' - four isomeric alcohols.