For this problem I suspect that you’re not intended to consider $0$ a natural number: if you include it, the natural recurrence breaks down between $n=1$ and $n=2$.
Each $n$-digit number is obtained from an $(n-1)$-digit number by multiplying it by $10$ and adding a one-digit integer (including $0$). Note that $10k\equiv k\pmod3$ (why?), so $10k+d\equiv k+d\pmod3$. For instance, $54=10\cdot 5+4\equiv5+4\equiv0\pmod 3$. So where do the members of $A_n$ come from? They are the numbers $10k+d$ such that $k$ is an $(n-1)$-digit number, $d$ is a digit, and $k+d\equiv0\pmod3$. That means that either $k\equiv d\equiv0\pmod3$, or $k\equiv 1\pmod3$ and $d\equiv2\pmod3$, or $k\equiv2\pmod3$ and $d\equiv1\pmod3$.
- If $k\equiv0\pmod3$, there are $4$ choices for $d$ — $0,3,6$, and $9$ — and there are $|A_{n-1}|$ choices for $k$; these account for $4|A_{n-1}|$ $n$-digit multiples of $3$.
- If $k\equiv1\pmod3$, there are $3$ choices for $d$ — $2,5$, and $8$ — and there are $|B_{n-1}|$ choices for $k$; these account for $3|B_{n-1}|$ $n$-digit multiples of $3$.
- And if $k\equiv2\pmod3$, there are $3$ choices for $d$ — $1,4$, and $7$ — and there are $|C_{n-1}|$ choices for $k$; these account for $3|C_{n-1}|$ $n$-digit multiples of $3$.
Thus, $|A_n|=4|A_{n-1}|+3|B_{n-1}|+3|C_{n-1}|$. Unfortunately, you’re asked to find $|A_n|$ just in terms of $|B_{n-1}|$ and $|C_{n-1}|$. Fortunately, it’s not too hard to get rid of the $A_{n-1}$ if you see that there are
$$(10^n-1)-(10^{n-1}-1)=10^n-10^{n-1}=9\cdot10^{n-1}$$
$n$-digit positive integers: that implies that $$|A_n|+|B_n|+|C_n|=9\cdot10^{n-1}$$ for each $n$. In particular, $|A_{n-1}|=9\cdot10^{n-2}-|B_{n-1}|-|C_{n-1}|$, so
$$\begin{align*}
|A_n|&=4|A_{n-1}|+3|B_{n-1}|+3|C_{n-1}|\\
&=4\big(9\cdot10^{n-2}-|B_{n-1}|-|C_{n-1}|\big)+3|B_{n-1}|+3|C_{n-1}|\\
&=36\cdot10^{n-2}-|B_{n-1}|-|C_{n-1}|\,.
\end{align*}$$
You can use the same kind of reasoning to get the other two recurrences.
I will note, however, that this is not really a very good problem, since it’s actually easier to derive the closed forms of $|A_n|,|B_n|$, and $|C_n|$ directly by observing that since $9\cdot10^{n-1}$ is a multiple of $3$, the $9\cdot10^{n-1}$ $n$-digit numbers can be divided into $3\cdot10^{n-1}$ blocks of $3$ consecutive integers, and each of these blocks contains one member of $A_n$, one member of $B_n$, and one member of $C_n$.
