In Cannizzaro reaction given below: the slowest step is :

Cannizzaro reaction is the base-induced disproportionation of an aldehyde having no α-hydrogen. Two molecules of the aldehyde convert, one to a carboxylate ion and the other to an alcohol.

Step-wise mechanism:
1. $$HO^-$$ adds to one molecule of the aldehyde to give an alkoxide (a tetrahedral intermediate).
2. That alkoxide acts as a hydride donor: it transfers $$H^-$$ to the carbonyl carbon of a second aldehyde molecule, producing a carboxylate ion and a new alkoxide (the alkoxide of the alcohol).
3. The new alkoxide is finally protonated by water to furnish the alcohol.

Kinetic studies give $$\text{rate}=k[\text{aldehyde}]^{2}[HO^-]$$. The rate depends on two aldehyde molecules and one hydroxide ion, so the slow (rate-determining) step must involve both aldehyde molecules after the first addition of $$HO^-$$. That is exactly the hydride-transfer step described in Step 2.

Nucleophilic addition of $$HO^-$$ (Step 1) and the final proton transfer (Step 3) are fast equilibria; the hydride migration (Step 2) requires breaking a C-H bond and forming a new C-H bond simultaneously, so it has the highest energy barrier and is the slowest.

Therefore, the slowest (rate-determining) step is the transfer of hydride to the carbonyl group.

Option B which is: the transfer of hydride to the carbonyl group