1. We need to form three-digit numbers using the digits $\{1,3,5,8\}$, with repetition allowed.

2. A number is divisible by $3$ if the sum of its digits is divisible by $3$.

3. First, classify the given digits by their remainders modulo $3$:

   • $1 \equiv 1 \pmod{3}$
   • $3 \equiv 0 \pmod{3}$
   • $5 \equiv 2 \pmod{3}$
   • $8 \equiv 2 \pmod{3}$

   So the available residue classes are:

   • remainder $0$: $\{3\}$
   • remainder $1$: $\{1\}$
   • remainder $2$: $\{5,8\}$

4. For a 3-digit number to be divisible by $3$, the sum of the three residues must be $0 \pmod{3}$.

   Possible residue patterns are:

   • $(0,0,0)$
   • $(1,1,1)$
   • one each of $(0,1,2)$
   • $(2,2,2)$ because $2+2+2=6 \equiv 0 \pmod{3}$

5. Count each case.

   Case 1: $(0,0,0)$

   • Only digit available is $3$.
   • Number formed: $333$
   • Count $=1$

   Case 2: $(1,1,1)$

   • Only digit available is $1$.
   • Number formed: $111$
   • Count $=1$

   Case 3: one each of $(0,1,2)$

   • Choose digits:
     • remainder $0$: $3$ (1 choice)
     • remainder $1$: $1$ (1 choice)
     • remainder $2$: either $5$ or $8$ (2 choices)
   • For each such selection, arrange the three distinct digits in $3! = 6$ ways.
   • Total count: $1 \cdot 1 \cdot 2 \cdot 6 = 12$

   Case 4: $(2,2,2)$

   • Each place can be filled by either $5$ or $8$.
   • Total such numbers: $2^3 = 8$

6. Add all counts: $1+1+12+8=22$

7. Therefore, the total number of required three-digit numbers is $\boxed{22}$

8. Comparing with the stored correct answer:

   • Stored answer: B
   • Our derived answer: 22, i.e. B

So the stored answer is correct.