Since the numbers are divisible by 5, in last place (unit's place)
we can have either 5 or 0
Fixing 0 in the last place we can have $5!$ =120 numbers.
Similarly,by fixing 5 at the last place we can have 120 numbers.
But out of these 120 numbers we have to eliminate
the numbers starting with 0.
Because these numbers will be 5 digit numbers.
$i.e.,$ Fixing 0 at the first place and 5 at the last place,
we can have $4!=24$ numbers which are 5 digit numbers.
$\therefore$ The required 6 digit numbers = $120+120-24=216$