The number of arrangements of the numbers strictly between 10 and 1000 formed with the digits 0, 1, 2, 3, 4, 5, 6 without repetition is

The numbers between $10$ and $1000$ consist of $2$-digit and $3$-digit numbers, strictly greater than $10$.

Available digits: $0, 1, 2, 3, 4, 5, 6$ (a total of $7$ digits).

For $2$-digit numbers:
The tens place can be filled in $6$ ways (excluding $0$).
The units place can be filled in $6$ ways (any of the remaining $6$ digits, including $0$).
Total $2$-digit numbers formed = $6 \times 6 = 36$.
Since the numbers must be strictly between $10$ and $1000$, the number $10$ itself must be excluded.
Valid $2$-digit numbers = $36 - 1 = 35$.

For $3$-digit numbers:
The hundreds place can be filled in $6$ ways (excluding $0$).
The tens place can be filled in $6$ ways.
The units place can be filled in $5$ ways.
Total $3$-digit numbers formed = $6 \times 6 \times 5 = 180$.

Total valid numbers = $35 + 180 = 215$.

Answer: $215$