1. Condition for divisibility by 3

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

We must form a 5-digit number using digits from $\{0,1,2,3,4,5\}$ without repetition.

So we choose any 5 digits out of these 6.

The sum of all six digits is $0+1+2+3+4+5=15$ which is divisible by $3$.

If one digit $d$ is omitted, then the sum of the chosen 5 digits is $15-d$ For divisibility by $3$: $15-d \equiv 0 \pmod{3}$ Since $15 \equiv 0 \pmod{3}$, we need $d \equiv 0 \pmod{3}$ So the omitted digit must be one of $0,3$.

Thus there are only 2 valid sets of 5 digits:

• omit $0$: digits $\{1,2,3,4,5\}$
• omit $3$: digits $\{0,1,2,4,5\}$

2. Count numbers for each valid set

Case 1: Omit $0$

Digits are $\{1,2,3,4,5\}$.

All 5 digits are nonzero, so every permutation gives a valid 5-digit number.

Number of such numbers: $5! = 120$

Case 2: Omit $3$

Digits are $\{0,1,2,4,5\}$.

Total permutations of these 5 digits: $5! = 120$

But those starting with $0$ are not 5-digit numbers. Fix $0$ in the first place, and arrange the remaining 4 digits: $4! = 24$

So valid 5-digit numbers in this case: $120-24=96$

3. Total count

Therefore, total numbers divisible by $3$ are $120+96=216$

4. Option check

The correct option is: $\boxed{216}$ So Option D is correct.