Given,

Three-digit number to be formed which are divisible by $3$ using the digits $1,3,5,8$ and repetition is allowed,

Now taking case $\left(1\right)$, where all digits are same, we get

$\left(1,1,1\right), \left(3,3,3\right), \left(5,5,5\right), \left(8,8,8\right)\to 4 \text{ways}$

Now taking case $\left(2\right)$, where $2-$digit are same and one is distinct, we get

$\left(5,5,8\right)\to \frac{3!}{2!}=3 \text{ways,} \left(8,8,5\right)\to 3 \text{ways}$

Now taking case $\left(3\right)$, where all are distinct, we get

$\left(1,3,5\right)\to 6\text{ ways}, \left(1,8,3\right)\to 6 \text{ways}$

So, total ways will be $4+3+3+6+6=22$