Answer:
Ac = 10100100011
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

Explanation:
All these operations are bitwise operations.
Ac is the complement of a. So where we have a bit 0, the complement is 1. Where we have a bit 1, the complement is 0. So
A = 01011011100
Ac = 10100100011
The second operation is the union between A and B. This is bitwise(bit 1 of A with bit 1 of B, bit 2 of A with bit 2 of B,...). The union operation is only 0 when it is between two zeros. So:
A = 01011011100
B = 10110111010

The third operation is the intersection between A and B. Again, bitwise. The intersection is only 1 when it is between two bits that are 1. So
A = 01011110100
B = 10110111010

The last operation is the bitwise subtraction between A and B. We start from the least significant bit(the last one). And we have to take care of the borrow operator also, similarly to a decimal subtraction.
We can only borrow from a previous bit 1, and this bit is set to 0
0-0 is 0 with no borrow
1-0 is 1 with no borrow
1 with borrow - 0 is 1 with borrow
1 with borrow - 1 is 1 with no borrow
0-1 is 1 with no borrow
0 with borrow -1 is 0 with no borrow
1-1 is 0 with no borrow
If we arrive at the first bit(the most significant) with a borrow, we must add a 1 at the front of the answer. So
A = 01011110100
B = 10110111010
