The van't Hoff factor ($i$) for a dilute aqueous solution of the strong electrolyte barium hydroxide is:

Barium hydroxide [${\text{Ba(OH)}}_2$ ] is a strong electrolyte and dissociates completely in aqueous solution:
$${\text{Ba(OH)}}_2\to {\text{Ba}}^{2+}+2{\text{OH}}^{-}$$
The total number of particles produced from one formula unit is $1({\text{Ba}}^{2+})+2({\text{OH}}^{-})=3$.Therefore, $i=3$.