No. of bit string of length 7 = $2^7 = 128$ as each position can be either 0 or 1.
Lets consider the ones with equal number of 0's and 1's. No such string is possible as string length is odd.
Now, both 1 and 0 are equally likely in a string. This means exactly half of 128 must have more number of 1's and same for more number of 0's. So, our answer must be 64.
Alternatively, we can say we require either 4 number of 1's or 5 number of 1's, or 6 number of 1's or 7 number of 1's each amounting to ${}^7C_4, {}^7C_5, {}^7C_6$ nd ${}^7C_7$ possibilities. So, 35 + 21 + 7 + 1 = 64.$