Here, first understand the problem.
$\mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_4=\{(a,b,c) \ | \ a \in \{0,1\},b \in \{0,1,2\},c \in \{0,1,2,3\} \}$
So, $\mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_4$ is a set of triplets in which first element comes from the set
$\{0,1\},$ second element comes from the set $\{0,1,2\}$ and third element comes from the set $\{0,1,2,3\}$.
It forms a group say $G$ under the operation say $\oplus$
such that:
$(a_1,b_1,c_1) \oplus (a_2,b_2,c_2) = (a_1 \oplus_2 a_2,b_1 \oplus_3 b_2,c_1 \oplus_4 c_2)$
Where $x \oplus_m y=(x+y) \ mod \ m$
Now, here, $x=x^{-1}$ means $x^2=e$ where $x$ and identity element $e$ belong to group $G$.
So, here we need to find the number of elements $x$ of $G$ such $x \oplus x = e$.
Now, for the group $G,$ the unique identity element $e=(0,0,0)$
because $(a,b,c) \oplus (0,0,0)=(a \oplus_2 0, b \oplus_3 0, c \oplus_3 0)=(a,b,c)$
Now, to find $x$ such that $x \oplus x = e$ i.e.
$(x_1,y_1,z_1) \oplus (x_1,y_1,z_1)= (x_1 \oplus_2 x_1,x_2 \oplus_3 x_2,x_3 \oplus_4 x_3)=(0,0,0)$
$(1)$ For $x_1$ we have $2$ choices that is $0$ and $1$ because $0 \oplus_2 0 =0$ and $1 \oplus_2 1 =0$
$(2)$ For $x_2$ we have $1$ choice that is $0$
because $0 \oplus_3 0 =0$ and for $x_2=1,2$
$1 \oplus_3 1 \neq 0$ and $2 \oplus_3 2 \neq 0$
$(3)$ For $x_3$ we have $2$ choices that is $0$ and $2$
because $0 \oplus_4 0 =0$ and $2 \oplus_4 2 =0$ for $x_2=1,3$
$1 \oplus_4 1 \neq 0$ and $3 \oplus_4 3 \neq 0$
Hence, for $x_1$ we have $2$ choices, for $x_2$ we have $1$ choice and for $x_3$ we have $2$ choices.
So, total choices such that $x^2=e$ is $4$.
Hence, Answer: 4
$\textbf{Brute Force}$
$\mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_4=$
$\{(0,0,0),(0,1,0),(0,2,0),(1,0,0),(1,1,0),(1,2,0),(0,0,1),(0,1,1),(0,2,1),(1,0,1),(1,1,1),$
$(1,2,1),(0,0,2),(0,1,2),(0,2,2),(1,0,2),(1,1,2),(1,2,2),(0,0,3),(0,1,3),(0,2,3),(1,0,3),(1,1,3),(1,2,3)\}$
Now, take each element from this set and check whether $x^2=e$ or not.
For example, $(0,0,0) \oplus (0,0,0) = (0 \oplus_2 0,0 \oplus_3 0,0 \oplus_4 0)=(0,0,0)$ here it is valid and similarly $(0,1,0) \oplus (0,1,0) = (0 \oplus_2 0,1 \oplus_3 1,0 \oplus_4 0)=(0,2,0) \neq (0,0,0)$ hence it is invalid.
When you do for all the $24$ elements of this set, you will get the valid elements as $(0,0,0),(1,0,0),(0,0,2)$ and $(1,0,2)$
Because
(i) $(0,0,0) \oplus (0,0,0) = (0 \oplus_2 0,0 \oplus_3 0,0 \oplus_4 0)=(0,0,0)$
(ii) $(1,0,0) \oplus (1,0,0) = (1 \oplus_2 1,0 \oplus_3 0,0 \oplus_4 0)=(0,0,0)$
(iii) $(0,0,2) \oplus (0,0,2) = (0 \oplus_2 0,0 \oplus_3 0,2 \oplus_4 2)=(0,0,0)$
(iv) $(1,0,2) \oplus (1,0,2) = (1 \oplus_2 1,0 \oplus_3 0,2 \oplus_4 2)=(0,0,0)$
