Given a $\sigma$-algebra $\mathcal{A}$ of subsets of $X$, a subset $\mathcal{B} \subset \mathcal{A}$ is a $\textbf{sub-algebra}$ of $\mathcal{A}$ if $\mathcal{B}$ is also a $\sigma$-algebra of subsets of $X$, i.e. $\emptyset \in \mathcal{B}$, $X \in \mathcal{B}$, and $\mathcal{B}$ is closed under complements and countable unions.
Some points to help think it through: if they call $(X, \mathcal{B})$ a measurable space, this can only mean $\mathcal{B}$ is a $\sigma$-alg. over $X$; also, remember that to define a $\sigma$-alg., you need to specify what set it is a $\sigma$-alg. over -- if they call $\mathcal{B}$ a sub-algebra of $\mathcal{A}$ without specifying what set $\mathcal{B}$ is a $\sigma$-alg. over, it is natural to conclude that it is over $X$.