Explanation:
To implement the Boolean function Y = AB CD using only 2-input NAND gates, we can follow the below steps:

Step 1: Draw the truth table for the given Boolean function.

A | B | C | D | Y
--|---|---|---|--
0 | 0 | 0 | 0 | 0
0 | 0 | 0 | 1 | 0
0 | 0 | 1 | 0 | 0
0 | 0 | 1 | 1 | 0
0 | 1 | 0 | 0 | 0
0 | 1 | 0 | 1 | 0
0 | 1 | 1 | 0 | 0
0 | 1 | 1 | 1 | 0
1 | 0 | 0 | 0 | 0
1 | 0 | 0 | 1 | 0
1 | 0 | 1 | 0 | 0
1 | 0 | 1 | 1 | 0
1 | 1 | 0 | 0 | 1
1 | 1 | 0 | 1 | 0
1 | 1 | 1 | 0 | 0
1 | 1 | 1 | 1 | 1

Step 2: From the truth table, we can see that the output Y is equal to 1 only when A = 1, B = 1, C = 0, and D = 0. So, we need to implement this condition using only 2-input NAND gates.

Step 3: We can use the De Morgan's theorem to simplify the expression for Y.

Y = AB CD
= (A' + B') + (C' + D')
= [(A' + B') (C' + D')]'

Step 4: We can implement the simplified expression using 3 NAND gates as shown below.

```
A B C D
| | | |
NAND NAND NAND NAND
| | | |
NAND--NAND--NAND--|
| | | |
| | | NAND
| | | |
| | NAND---NAND
| | | |
NAND---NAND---NAND--|
| | | |
| | | NAND
| | | |
| | NAND---NAND
| | | |
NAND---NAND--NAND--|
| | | |
| | | NAND
| | | |
NAND---NAND---NAND--Y
```

Therefore, the minimum number of gates required to realize the Boolean function Y = AB CD using only 2-input NAND gates is 3. Hence, option B is the correct answer.