1. Treat girls and boys as two blocks

Since all 3 girls must stand together and all 4 boys must stand together, we form:

• one block of 3 girls: $G$
• one block of 4 boys: $B$

These two blocks can be arranged in: $2! = 2$ ways.

2. Arrange the girls within their block

The 3 girls can be arranged among themselves in: $3! = 6$ ways.

3. Arrange the boys within their block with restriction

There are 4 boys total, including $B_1$ and $B_2$.

Total arrangements of 4 boys: $4! = 24$

Now count arrangements where $B_1$ and $B_2$ are adjacent. Treat $B_1$ and $B_2$ as one unit. Then we have 3 units:

• $(B_1B_2)$ or $(B_2B_1)$
• boy 3
• boy 4

These 3 units can be arranged in: $3! = 6$ ways, and inside the pair, $B_1,B_2$ can be arranged in: $2! = 2$ ways.

So, number of arrangements with $B_1$ and $B_2$ adjacent is: $3! \cdot 2! = 6 \cdot 2 = 12$

Therefore, number of boy arrangements where $B_1$ and $B_2$ are not adjacent: $4! - 12 = 24 - 12 = 12$

4. Total number of required arrangements

Multiply all independent choices: $2! \cdot 3! \cdot 12 = 2 \cdot 6 \cdot 12 = 144$

5. Compare with options

Thus the required number of ways is: $\boxed{144}$

So the correct option is D.