The question is how many equivalence relation can be defined on a set of $5$?
I think this is asking how many different ways can we partition a set of $5$, right?
So the answer is
$1$ way: $$\{1\},\{2\},\{3\},\{4\},\{5\}$$
$_5C_2$ ways: $$\{1,2\},\{3\},\{4\},\{5\}$$
$_5C_2\times _3C_2$ ways: $$\{1,2\},\{3,4\},\{5\}$$
$_5C_3$ ways: $$\{1,2,3\},\{4\},\{5\}$$
$_5C_3$ ways: $$\{1,2,3\},\{4,5\}$$
$_5C_4$ ways: $$\{1,2,3,4\},\{5\}$$
$1$ way: $$\{1,2,3,4,5\}$$
So the total number is $1+10+30+10+10+5+1=67$.
Is that correct?
You've made a calculation error, you have double-counted the partitions of type $\{a,b\},\{c,d\},\{e\}$, since $\{a,b\},\{c,d\},\{e\}$ is the same partition as $\{c,d\},\{a,b\},\{e\}$. There are only $15$ of those, not $30$. The correct number of partitions (therefore also the correct number of equivalence classes) is $52$, the $5$th Bell number.