Let A denote the event that a 6 -digit integer formed by 0, 1, 2,...

Total cases : $6 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2$

$n\left(s\right) = 6 \cdot 6 !$

Favourable cases :

Number divisible by $3 ≡$ Sum of digits must be divisible by $3$

Case-$I$

$1, 2, 3, 4, 5, 6$

Number of ways $= 6 !$

Case-$II$

$0, 1, 2, 4, 5, 6$

Number of ways $= 5 \cdot 5 !$

Case-$III$

$0, 1, 2, 3, 4, 5$

Number of ways $= 5 \cdot 5 !$

$n ($ favourable $) = 6 ! + 2 \cdot 5 \cdot 5 !$

$P = \dfrac{6 ! + 2 \cdot 5 \cdot 5 !}{6 \cdot 6 !} = \dfrac{4}{9}$