If the numerical value of a 2-byte unsigned integer on a little endian computer is 255 more than that on a big endian computer, which of the following choices represent(s) the unsigned integer on a little endian computer?

2,3 is the correct answer. 

concept:

If you split hex number into bytes,
In hexadecimal representation, each digit corresponds to 4 binary digits.(i.e 2 hex digits = 8 bits= 1 byte). 

If  little-endian representation used then byte arrangement is as follows:

if Big endian representation is used then the byte arrangement is as follows:

Within the byte no change in the bit sequence.

A) 0x0001 is incorrect.
Little-endian 0x0001  gives you 01 as the first byte and 00 as the second byte, So the corresponding big-endian representation is 0x0100.
Hex (0x0001) is equal to 1 in decimal 
Hex(0x0100) is equal to 256 in decimal. 
Their difference is -255.

B) 0x6665 is correct
Little-endian 0x6665  gives you 65 as the first byte and 66 as the second byte, So the corresponding big-endian representation is 0x6566.
Hex (0x6665) is equal to 26213 in decimal 
Hex(0x6566) is equal to 25958 in decimal. 
Their difference is 255.

C) 0x0100 is correct
Little-endian 0x0100  gives you 00 as the first byte and 01 as the second byte, So the corresponding big-endian representation is 0x0100
Hex(0x0100) is equal to 256 in decimal.,
Hex (0x0001) is equal to 1 in decimal 

Their difference is 255.

D)0x4243 is incorrect
Little-endian 0x4243  gives you 43 as first byte and 42 as the second byte, So the corresponding big-endian representation is 0x4342
Hex(0x4243) is equal to 16963  in decimal.
Hex (0x4342) is equal to  17218 in decimal. 
Their difference is -255.

So 2,3 is the answer