Solution:
Size of display $=400\times 160 \times 3 \; \text{Bytes} = 192,000 \; \text{Bytes} $
So, at $10 \;\text{fps}$ it will be $=1,920,000 \;\text{Bytes per second} = 15,360, 000 \;\frac{\text{bits}}{\text{sec}}$
Hence, it will consume $15\%$ of the Fast Ethernet.