Final answer:
The size of a Color Lookup Table (LUT) given a 10-bit frame buffer with RGB components having 8-bits each is calculated based on the 2^10 pixel values and the 24-bit full color information needed per entry.
Step-by-step explanation:
If the frame buffer has 10-bits per pixel and 8-bits are allocated for each of the r,g, and b components, to calculate the size of the Color Lookup Table (LUT), you need to consider how many different colors can be represented by those bits. Each RGB component can represent 28 different values (from 0 to 255), but since you only have 10 bits for the pixel in total, the size of the LUT will be based on the number of different pixel values that those 10 bits can represent, which is 210.
Therefore, the size of the LUT will be 210 entries. However, since each LUT entry typically needs to store the full 24-bit color (8 bits for red, 8 bits for green, 8 bits for blue), the total size of the LUT in bytes will be 210 entries times 24 bits per entry. First, convert 24 bits to bytes by dividing by 8 (since there are 8 bits in a byte), which gives us 3 bytes per entry.
Finally, 210 entries times 3 bytes per entry equals 210 * 3 bytes, or 3072 bytes, which is not one of the options provided. Therefore, the question might contain a mistake, or it may be considering a different architecture for the Color Lookup Table (LUT).