Since I have the time, here is a pretty thorough explanation of how you might attack the problem by brute force, followed by some pointers on how the amount of brute force could be reduced.
Since $\{1,2,3,4,5\}$ has $5$ elements, it has $2^5=32$ subsets. That’s a small enough number that if you can’t find a nicer approach, you can simply list them all, find their intersections with $\{1,3,5\}$, and use that to sort them into equivalence classes.
$\begin{array}{c} \text{Set}&\text{Intersection with }\{1,3,5\}\\ \hline \varnothing&\varnothing\\ \{1\}&\{1\}\\ \{2\}&\varnothing\\ \{3\}&\{3\}\\ \{4\}&\varnothing\\ \{5\}&\{5\}\\ \{1,2\}&\{1\}\\ \{1,3\}&\{1,3\}\\ \{1,4\}&\{1\}\\ \color{red}{\{1,5\}}&\{1,5\}\\ \{2,3\}&\{3\}\\ \{2,4\}&\varnothing\\ \{2,5\}&\{5\}\\ \{3,4\}&\{3\}\\ \{3,5\}&\{3,5\}\\ \{4,5\}&\{5\}\\ \{1,2,3\}&\{1,3\}\\ \{1,2,4\}&\{1\}\\ \color{red}{\{1,2,5\}}&\{1,5\}\\ \{1,3,4\}&\{1,3\}\\ \{1,3,5\}&\{1,3,5\}\\ \color{red}{\{1,4,5\}}&\{1,5\}\\ \{2,3,4\}&\{3\}\\ \{2,3,5\}&\{3,5\}\\ \{2,4,5\}&\{5\}\\ \{3,4,5\}&\{3,5\}\\ \{1,2,3,4\}&\{1,3\}\\ \{1,2,3,5\}&\{1,3,5\}\\ \color{red}{\{1,2,4,5\}}&\{1,5\}\\ \{1,3,4,5\}&\{1,3,5\}\\ \{2,3,4,5\}&\{3,5\}\\ \{1,2,3,4,5\}&\{1,3,5\} \end{array}$
As an example, the sets shown in are the sets that have $\{1,5\}$ as their intersection with $\{1,3,5\}$, so they form one equivalence class of the relation.
Very likely you would realize part of the way through that it would be simpler to start with the possible intersections with $\{1,3,5\}$ and systematically list the subsets of $\{1,2,3,4,5\}$ having each possible intersection, like this:
$\begin{array}{c|l} \text{Intersection with }\{1,3,5\}&\text{Subsets having that intersection}\\ \hline \varnothing&\varnothing,\{2\},\{4\},\{2,4\}\\ \{1\}&\{1\},\{1,2\},\{1,4\},\{1,2,4\}\\ \{3\}&\{3\},\{2,3\},\{3,4\},\{2,3,4\}\\ \{5\}&\{5\},\{2,5\},\{4,5\},\{2,4,5\}\\ \{1,3\}&\{1,3\},\{1,2,3\},\{1,3,4\},\{1,2,3,4\}\\ \{1,5\}&\color{red}{\{1,5\},\{1,2,5\},\{1,4,5\},\{1,2,4,5\}}\\ \{3,5\}&\{3,5\},\{2,3,5\},\{3,4,5\},\{2,3,4,5\}\\ \{1,3,5\}&\{1,3,5\},\{1,2,3,5\},\{1,3,4,5\},\{1,2,3,4,5\} \end{array}$
In constructing this table you should notice that each row can be obtained from the top row in a very simple way. For example, the equivalence class of $\{1,5\}$, again in red, is obtained by taking the union of $\{1,5\}$ with each of the sets having empty intersection with $\{1,3,5\}$. You should think about why this must be the case: if $A\cap\{1,3,5\}=\{1,5\}$, then it must be possible to split $A$ into two disjoint pieces, $\{1,5\}$ and the set $A'$ of elements of $A$ that are not in $\{1,5\}$. This $A'$ is disjoint from $\{1,3,5\}$, so it must be one of the four sets having intersection $\varnothing$ with $\{1,3,5\}$. Thus, the red line is just
$\{1,5\}\cup\varnothing,~\{1,5\}\cup\{2\},~\{1,5\}\cup\{4\},~\{1,5\}\cup\{2,4\}\;,$
and each of the other lines below the first is obtained similarly.
Now recall that $x\sim y$ if and only if $x\cap\{1,3,5\}=y\cap\{1,3,5\}$. The four sets shown in red all have the same intersection with $\{1,3,5\}$, so they are all related: $\{1,5\}\sim\{1,4,5\}$, $\{1,5\}\sim\{1,5\}$, $\{1,2,4,5\}\sim\{1,4,5\}$, etc. The collection $\Big\{\{1,5\},\{1,2,5\},\{1,4,5\},\{1,2,4,5\}\Big\}$ is one equivalence class of the relation $\sim$; all of them are shown in the table above. As you can see, there are eight of them, each with four elements.
Having done everything else, I’ll leave it to you to identify the equivalence class to which $\{1,2,3\}$ belongs.