We define the Van't Hoff factor as the number of particles each solute formula unit breaks apart into when it dissolves.
Here we are given $\mathit{N}{\mathit{a}}_2\mathit{S}{\mathit{O}}_{4.10}{\mathit{H}}_2\mathit{O}$
The water of crystallization which $10{\mathit{H}}_2\mathit{O}$ will not be counted in the dissociation number.
So, dissociation of a dilute solution of $\mathit{N}\mathit{a}2\mathit{S}{\mathit{O}}_4$ will be given as
$\mathit{N}\mathit{a}2\mathit{S}{\mathit{O}}_4\rightleftharpoons 2\mathit{n}\mathit{a}++\mathit{S}{\mathit{O}}_4^{2-}$
In this chemical equation, we can see that there are $2\mathit{N}\mathit{a}+$ and $1\mathit{S}{\mathit{O}}_4^{2-}$ ions. That makes the total number of ions equal to 3.
So, that means Van’t Hoff factor for $\mathit{N}{\mathit{a}}_2\mathit{S}{\mathit{O}}_{4.10}{\mathit{H}}_2\mathit{O}$ is also 3.