The gas that can be dried using concentrated tetraoxosulphate(VI) acid ($\text{H}_2\text{SO}_4$) must not react with it. Concentrated $\text{H}_2\text{SO}_4$ is a strong acid, an oxidizing agent, and a dehydrating agent. Let's examine each option: A) Hydrogen bromide ($\text{HBr}$): $\text{HBr}$ is a reducing agent. Concentrated $\text{H}_2\text{SO}_4$ will oxidize $\text{HBr}$ to bromine ($\text{Br}_2$). $$ \text{2HBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{SO}_2 + \text{2H}_2\text{O} $$ Therefore, $\text{HBr}$ cannot be dried by concentrated $\text{H}_2\text{SO}_4$. B) Sulphur(IV) oxide ($\text{SO}_2$): $\text{SO}_2$ is an acidic gas. It does not react with concentrated $\text{H}_2\text{SO}_4$ because both are acidic. Therefore, $\text{SO}_2$ can be dried by concentrated $\text{H}_2\text{SO}_4$. C) Hydrogen sulphide ($\text{H}_2\text{S}$): $\text{H}_2\text{S}$ is a strong reducing agent. Concentrated $\text{H}_2\text{SO}_4$ will oxidize $\text{H}_2\text{S}$ to sulfur ($\text{S}$). $$ \text{H}_2\text{S} + \text{H}_2\text{SO}_4 \rightarrow \text{S} + \text{SO}_2 + \text{2H}_2\text{O} $$ Therefore, $\text{H}_2\text{S}$ cannot be dried by concentrated $\text{H}_2\text{SO}_4$. D) Ammonia ($\text{NH}_3$): $\text{NH}_3$ is a basic gas. It will react with the acidic concentrated $\text{H}_2\text{SO}_4$ to form ammonium sulfate. $$ \text{2NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4 $$ Therefore, $\text{NH}_3$ cannot be dried by concentrated $\text{H}_2\text{SO}_4$. Based on this analysis, only sulphur(IV) oxide can be dried using concentrated tetraoxosulphate(VI) acid. The final answer is $\boxed{\text{B}}$.