Identify substrate ' $A$ ' in the following reaction. $\mathrm{A}+\mathrm{CO}_2 \xrightarrow[6 \mathrm{~atm}]{398 \mathrm{~K}}$ Sodium salicylate $\xrightarrow{\mathrm{H}_3 \mathrm{O}^{+}}$Salicylic acid

Kolbe-Schmitt reaction analysis:

Sodium phenoxide carboxylation with $\mathrm{CO_2}$ at 398 K under 6 atm pressure yields sodium salicylate, which upon acidification produces salicylic acid:

$\mathrm{C_6H_5ONa} + \mathrm{CO_2} \xrightarrow[]{398 \mathrm{~K}, 6 \mathrm{~atm}} \mathrm{o-HOC_6H_4COONa}$

$\mathrm{o-HOC_6H_4COONa} + \mathrm{H_3O^+} \longrightarrow \mathrm{o-HOC_6H_4COOH}$

Given the reaction conditions and products, the substrate A must be sodium phenoxide.

Final answer: $\boxed{\text{B}}$