Question: I can't seem to wrap my head around this problem.How do I make sure that the digit 0 and digits from 7 to 9 are not included ? Provided- Repetition of digits is allowed.

Answer: 
To be a multiple of $3$, the sum of the three digits must be a multiple of $3$.
After choosing the first two digits freely, that specifies a congruence class
modulo $3$ for the last digit, and there are two possibilities within that range.
For instance, starting with $52$, the final digit must be $2$ or $5$, etc.
Let be $A = \{ 0, 1, 2 \}$ least residue system modulo 3.  

Let be 
$$
 B\_1 = \{ (a, b, c) \in A\times A\times A : a + b + c = 3\}
$$
$B\_1$ has $\frac{5!}{3!2!} - 3 = 7$ elements, then $2^3 \cdot 7 = 56$ numbers with digits $1\dots6$ such that the sun of these digits modulo 3 is 3.  

At last, 
$$
 B\_2 = \{ (a, b, c) \in A\times A\times A : a + b + c = 6 \text{ or } a + b + c = 0 \}
$$
$B\_2 = \{ (0, 0, 0), (2, 2, 2) \}$, then, $16$ numbers with this property.  

Hence, They are $72$ numbers.