We have

$$ R=\{(x,y)\in \mathbb{N}\times \mathbb{N}:\ln(x+y)\le 2\}. $$

To make the relation transitive, we first find all pairs already in $R$.

Since

$$ \ln(x+y)\le 2 \iff x+y\le e^2. $$

Now,

$$ e^2 \approx 7.389. $$

So, for natural numbers $x,y$, we must have

$$ x+y\le 7. $$

Hence,

$$ R=\{(x,y)\in \mathbb{N}\times \mathbb{N}:x+y\le 7\}. $$

Now list all such pairs:

For $x=1$: $(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)$
For $x=2$: $(2,1),(2,2),(2,3),(2,4),(2,5)$
For $x=3$: $(3,1),(3,2),(3,3),(3,4)$
For $x=4$: $(4,1),(4,2),(4,3)$
For $x=5$: $(5,1),(5,2)$
For $x=6$: $(6,1)$

So these are the elements of $R$.

A relation is transitive if whenever

$$ (a,b)\in R \quad \text{and} \quad (b,c)\in R, $$

then

$$ (a,c)\in R. $$

We need to find all pairs $(a,c)$ that must be present because of transitivity.

Step 1: Find which middle elements are possible

If $(a,b)\in R$, then

$$ a+b\le 7. $$

If $(b,c)\in R$, then

$$ b+c\le 7. $$

For a fixed $b$, this gives

$$ a\le 7-b,\qquad c\le 7-b. $$

So for each $b$, transitivity demands that all pairs

$$ (a,c)\quad \text{with } 1\le a\le 7-b,\; 1\le c\le 7-b $$

must be in the relation.

Taking union over all possible $b\in \mathbb{N}$ with at least one such pair, we get:

For $b=1$: required pairs are all

$$ (a,c),\quad 1\le a\le 6,\; 1\le c\le 6. $$

This is already the largest possible set forced by transitivity.

So, if relation is to be transitive, it must contain every pair

$$ (a,c)\in \{1,2,3,4,5,6\}\times \{1,2,3,4,5,6\}. $$

That means the transitive closure must at least contain all $36$ pairs

$$ (a,c),\quad 1\le a,c\le 6. $$

Step 2: Count how many of these are already in $R$

Among these $36$ pairs, $R$ already contains those with

$$ a+c\le 7. $$

Count them:

$a=1$: $c=1$ to $6$ $\Rightarrow 6$
$a=2$: $c=1$ to $5$ $\Rightarrow 5$
$a=3$: $c=1$ to $4$ $\Rightarrow 4$
$a=4$: $c=1$ to $3$ $\Rightarrow 3$
$a=5$: $c=1$ to $2$ $\Rightarrow 2$
$a=6$: $c=1$ $\Rightarrow 1$

Total:

$$ 6+5+4+3+2+1=21. $$

So the number of additional pairs needed is

$$ 36-21=15. $$

Step 3: Check that this is sufficient

If we add all missing pairs among

$$ \{1,2,3,4,5,6\}\times \{1,2,3,4,5,6\}, $$

then the relation becomes

$$ \{(a,c):1\le a\le 6,\;1\le c\le 6\}. $$

This is clearly transitive, because whenever

$$ (a,b)\in R \quad \text{and} \quad (b,c)\in R, $$

all three numbers lie in $\{1,2,3,4,5,6\}$, so $(a,c)$ is also in the relation.

Therefore, the minimum number of elements to be added is

$$ \boxed{15} $$

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