The number of carbon atoms present in the product B is

Step 1: Identify product A from the fermentation of glucose.

$$\text{C}_6\text{H}_{12}\text{O}_6 \xrightarrow{\text{Zymase}} 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2$$

Product A is ethanol ($$\text{C}_2\text{H}_5\text{OH}$$).

Step 2: Identify product B from the iodoform reaction.

NaOI is sodium hypoiodite, the reagent for the iodoform reaction. Ethanol undergoes the iodoform reaction:

$$\text{CH}_3\text{CH}_2\text{OH} + 4\text{NaOI} \xrightarrow{\Delta} \text{CHI}_3 + \text{HCOONa} + 3\text{NaOH}$$

The products are iodoform ($$\text{CHI}_3$$) and sodium formate ($$\text{HCOONa}$$).

Therefore, product B is sodium formate ($$\text{HCOONa}$$), which contains 1 carbon atom.

The number of carbon atoms in product B is 1.