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.
When fixing two of the digits i.e. hundreds and tens, we get exactly two numbers from $1-6$ which placed at ones digit will make the number divisible by $3$. You can prove this by observing that there are exactly $3$ numbers of the form $3n,3n+1,3n+2$ in $1-6$.  

You are free to chose first two digits then you can have your last digit in two numbers.

So the number of numbers $=$ $6\times6\times2=72$