1. Let the total distance be $D$.

2. First half of the distance

   The particle covers distance $\dfrac{D}{2}$ with speed $6\,\text{m/s}$.

   Time taken: $t_1=\frac{D/2}{6}=\frac{D}{12}$

3. Second half of the distance

   This half is covered in two equal time intervals with speeds $9\,\text{m/s}$ and $15\,\text{m/s}$.

   Let each time interval be $t$.

   Then distance covered in these two intervals is: $9t+15t=24t$

   But this equals the remaining half distance: $24t=\frac{D}{2}$ $t=\frac{D}{48}$

   So total time for the second half is: $t_2=2t=\frac{D}{24}$

4. Total time of motion $T=t_1+t_2=\frac{D}{12}+\frac{D}{24}=\frac{2D+D}{24}=\frac{3D}{24}=\frac{D}{8}$

5. Average speed $v_{\text{avg}}=\frac{\text{total distance}}{\text{total time}}=\frac{D}{D/8}=8\,\text{m/s}$

6. Option check

   • A: $9.2\,\text{m/s}$ ❌
   • B: $8.8\,\text{m/s}$ ❌
   • C: $10\,\text{m/s}$ ❌
   • D: $8\,\text{m/s}$ ✅

Therefore, the average speed is $\boxed{8\,\text{m/s}}$