1. Identify compound A formed by zymase action

Zymase converts glucose by alcoholic fermentation:

$C_6H_{12}O_6 \xrightarrow{\text{zymase}} 2C_2H_5OH + 2CO_2$

So, compound A = ethanol $\left(C_2H_5OH\right)$.

2. Reaction of ethanol with $\text{NaOI}$ and heat

$\text{NaOI}$ is sodium hypoiodite, which gives the iodoform reaction.

Ethanol is first oxidized to acetaldehyde, and then undergoes haloform cleavage:

$CH_3CH_2OH \xrightarrow{NaOI,\,\Delta} HCOONa + CHI_3$

Thus, compound B = sodium formate $\left(HCOONa\right)$.

3. Count carbon atoms in B

In sodium formate, $HCOONa$, there is only one carbon atom.

4. Final answer

$\boxed{1}$

5. Comparison with stored answer

Stored correct answer = $1$.

My derived answer also = $1$, so they agree.