1. Divisibility rule for 11

For a 6-digit number with digits $d_1d_2d_3d_4d_5d_6$, the number is divisible by $11$ if

is a multiple of $11$.

Here the digits used are exactly $0,1,2,5,7,9$ once each.

2. Total sum of all digits

Let the sum of digits in odd positions be $S_o$ and in even positions be $S_e$. Then

and divisibility by $11$ requires

Since both $S_o,S_e$ are sums of three of the given digits, the difference cannot be $\pm 22$. Also, because $24$ is even, $S_o-S_e$ must be even, so it cannot be $\pm 11$. Hence the only possibility is

So we need to split the 6 digits into two groups of 3 each, each summing to $12$.

3. Find 3-digit subsets with sum 12

From $\{0,1,2,5,7,9\}$, the 3-element subsets summing to $12$ are:

• $\{0,5,7\}$
• $\{1,2,9\}$

These are complementary, so one must occupy the odd positions and the other the even positions.

Thus there are 2 ways to choose which set goes to odd positions:

• odd positions: $\{0,5,7\}$, even positions: $\{1,2,9\}$
• odd positions: $\{1,2,9\}$, even positions: $\{0,5,7\}$

4. Arrange digits in positions, respecting first digit nonzero

We now count arrangements for each case.

Case 1: Odd positions get $\{0,5,7\}$, even positions get $\{1,2,9\}$

Odd positions are $1,3,5$. Since the first digit cannot be $0$, among the $3! = 6$ arrangements of $\{0,5,7\}$ in odd positions, those with $0$ at position 1 are invalid.

If $0$ is fixed at position 1, the remaining two digits can be arranged in $2! = 2$ ways. So valid arrangements for odd positions:

Even positions can be filled with $\{1,2,9\}$ in

ways.

Hence total for Case 1:

Case 2: Odd positions get $\{1,2,9\}$, even positions get $\{0,5,7\}$

Now the first digit is one of $1,2,9$, so no restriction from leading zero.

Odd positions can be arranged in

ways.

Even positions can be arranged in

ways.

Hence total for Case 2:

5. Total count

Therefore, the number of such 6-digit numbers is

So the correct option is B.