Let's take R as:
R: {(1,1) (2,2) (3,3) (1,2) (2,1) (2,3) (1,3)} R is reflexive and transitive.
Now by the definition of E in the question,
E: {(1,1) (2,2) (3,3) (1,2) (2,1)}
Now, Diagonal elements are all present => E is reflexive
For symmetric, if every pair aRb, the pair bRa should be present. True.=> E is symmetric
For transitive, if pairs aRb and bRc are present, pair aRc should be present. True. =>E is transitive.
==>E is an equivalence relation.
Coming to the second part, the equivalence classes of E are:
[1] : {1,2} (let E1)
[2]: {1,2} (let E2)
[3]: {3} (let E3)
For the relation ≤ defined on equivalence classes of E,
For reflexivity, E1 ≤ E1
E2 ≤ E2
and E3 ≤ E3 should hold. For each of these we can find some "a∈E1,b∈E2 and (a,b)∈R" such that it holds.
eg. for E1 ≤ E1, a=1, b=1 and (1,1) ∈ R
For antisymmetric: E1 ≤ E2 and E2 ≤ E1 only if E1 = E2
We can see that E1 ≤ E2 and E2 ≤ E1 both hold and E1 and E2 are indeed equal. (Both equal {1,2}).
For transitivity, if E1 ≤ E2 and E2 ≤ E3 then E1 ≤ E3 should hold.
Here E1 ≤ E2 holds but E2 ≤ E3 does not, so we need not check for E2 ≤ E3. This makes the relation ≤ already transitive.
As the relation ≤ is reflexive, antisymmetric and transitive, it follows that it is a partial order.