$(\mathbb{Q},\times)$ is an algebraic structure because $\mathbb{Q}$ is closed under multiplication operation.
It is a semi group because $'\times'$ is associative on $\mathbb{Q}$.
It is a monoid because identity element $e=1\in \mathbb{Q}$.
A monoid $(\mathbb{Q},\times)$ with identity element $'e'$ is called a group if for each element $a\in \mathbb{Q},$ there exists an element $b\in \mathbb{Q},$ such that $(a \times b) = (b \times a) = e.$
Then $'b'$ is called inverse of an element $'a',$ denoted by $a^{-1}$.
Now, $a\times b = 1$
$\implies \boxed{b = \dfrac{1}{a}}$
Inverse of $'a'$ is $\dfrac{1}{a}$
Inverse of $'0'$ is $\dfrac{1}{0}\notin \mathbb{Q}$
It is not a group.
So, the correct answer is $(B)$. For $A, C$ and $D,$ both identity as well as inverse exists.