Ans: A
Hamming Distance:- You have to do XOR between each Code , then count number of 1’s in output
$\Rightarrow$ Hamming distance between 00000 and 01011
00000 ⊕ 01011 = 01011 no. of 1’s in(01011) = 3 i:e 3
$\Rightarrow$ Hamming distance between 01011 and 10101
01011 ⊕ 10101 = 11110 no. of 1’s in(11110) = 4 i:e 4
if you do for each code then you will find that minimum hamming distance = 3
Maximum number of erroneous bits that can be corrected by the code = d
(2*d + 1) = Minimum Hamming distance
2*d + 1 = 3
d = 1