Let R be an equivalence relation on the set A = {1, 2, 3, 4, 5} given by R = {(x, y):2 divides (x - y)}. Then equivalence class of 3 is:

The correct option is {1, 3, 5} (written as (1, 3, 5} in the options).

To find the equivalence class of 3, denoted by [3], we need to find all elements $x\in A$ such that $(x,3)\in R$.

According to the definition of the relation R:
$(x,y)\in R\Rightarrow 2\text{divides}(x-y)$

This means that $(x,3)\in R$ if and only if 2 divides $(x-3)$.
For 2 to divide $(x-3)$, the difference $(x-3)$ must be an even integer.

Let us test each element $x$ of the set A = {1, 2, 3, 4, 5}:

For $x=1$:
$x-3=1-3=-2$
Since -2 is divisible by 2, $1\in \left[3\right]$.

For $x=2$:
$x-3=2-3=-1$
Since -1 is not divisible by 2, $2\notin \left[3\right]$.

For $x=3$:
$x-3=3-3=0$
Since 0 is divisible by 2, $3\in \left[3\right]$.

For $x=4$:
$x-3=4-3=1$
Since 1 is not divisible by 2, $4\notin \left[3\right]$.

For $x=5$:
$x-3=5-3=2$
Since 2 is divisible by 2, $5\in \left[3\right]$.

Gathering all elements that belong to the equivalence class of 3, we get:
$\left[3\right]=\{1,3,5\}$