1. Interpret the sets

• $A=$ set of all 3-digit natural numbers, so $A=\{100,101,102,\dots,999\}$
• $B=\{9k+2:k\in \mathbb N\}$, i.e. numbers congruent to $2 \pmod 9$.
• $C=\{9k+l:k\in \mathbb N\}$, i.e. numbers congruent to $l \pmod 9$, where $0<l<9$.

We need the sum of all elements of $A\cap (B\cup C)$ to be $274\times 400=109600.$

So among the 3-digit numbers, we include those congruent to either $2$ or $l$ modulo $9$.

2. Count and sum 3-digit numbers in each residue class mod 9

Since $100\equiv 1 \pmod 9, \qquad 999\equiv 0 \pmod 9,$ the 900 three-digit numbers split equally among the 9 residue classes modulo 9.

Hence each residue class has $\frac{900}{9}=100$ terms.

Also, numbers in any fixed residue class form an arithmetic progression with common difference $9$.

3. Sum of 3-digit numbers congruent to $2 \pmod 9$

The first 3-digit number congruent to $2 \pmod 9$ is $101$ and the last is $992$.

So the progression is $101,110,119,\dots,992$ with $100$ terms.

Its sum is $S_2=\frac{100}{2}(101+992)=50\cdot 1093=54650.$

4. Required sum for residue class $l$

If $l\neq 2$, then the two residue classes are disjoint, so $S_2+S_l=109600.$ Thus $S_l=109600-54650=54950.$

Now compute $S_l$ in terms of $l$.

For residue class $l$, the smallest 3-digit number congruent to $l \pmod 9$ is:

• if $l\ge 1$, starting from $100\equiv 1 \pmod 9$, the first such number is $99+l$.

Indeed, $99+l \equiv l \pmod 9,$ and since $1\le l\le 8$, this lies between $100$ and $107$.

The largest 3-digit number congruent to $l \pmod 9$ is $990+l.$

So the AP is $(99+l),(108+l),(117+l),\dots,(990+l)$ with $100$ terms.

Hence $S_l=\frac{100}{2}\big((99+l)+(990+l)\big)$ $=50(1089+2l).$

Set this equal to $54950$: $50(1089+2l)=54950$ $1089+2l=1099$ $2l=10$ $l=5.$

5. Check the case $l=2$

If $l=2$, then $B=C$, so the sum would just be $S_2=54650$, not $109600$. Thus $l\neq 2$, and our solution is valid.

6. Final answer

$\boxed{5}$