$\underline{\textbf{Answer:}\Rightarrow}\;\textbf{(B)}\;10$
$\underline{\textbf{Explanation:}\Rightarrow}$
$\underline{\mathbf{Method:\; 1}\Rightarrow}$
Round Trip propogation delay $=40\;\text{ms}$
Frame Size $=32\times 8\;\text{bits}$
Transmission Time $=\mathbf{\dfrac{L}{B}} = \dfrac{32 \times 8}{64} \;\text{ms} = 4\;\text{ms}$
Let $\mathbf n$ be the window size.
$\text{Utilizaton} =\mathbf{\dfrac{n}{1+2a}}$, where $\mathrm {\mathbf a = \dfrac{Propogation\; time}{transmission\; time}} = \dfrac{\mathrm n}{1+\dfrac{40}{4}}$
For maximum Utilization, Efficiency $ = 1$
$\Rightarrow 1 = \dfrac{\mathrm n}{1+\dfrac{40}{4}}\\ \Rightarrow\mathrm n = 11$
Which is close to option $\mathbf B$
Or, we can do it another way as well. Like ignoring $\mathbf 1$ from $\mathbf{1+2a}$ which few authors do as well.
$\therefore\;\mathbf B$ is the correct answer.
$\underline{\textbf{Method:}\; \mathbf 2\Rightarrow}$
Round Trip propagation delay $=40\;\text{ms}$
Size of frame $ = 32\times 8$
Bandwidth $=64\;\text{kbps}$
Total Data $= 40 \times 64\;\text{bits} = 40 \times \dfrac{64}{8}\;\text{Bytes}= 320 \;\text{Bytes}$
Size of single packet $ = 32 \; \text{Bytes}$
Number of packets $= \dfrac{\text{Total Data}}{\text{Frame Size}}=\dfrac{320}{32} = 10 \;\text{Packets}$
$\therefore \mathbf{B}$ is the correct option.