Hamming code can correct 1 bit error. So if more than one bit gets flipped decoding will result in wrong codeword.
So P(more than 1 bit gets flipped) = 1 - P(0 or 1 bit gets flipped) = 1 - P(no bit gets flipped) - P(1 bit gets flipped)
P(particular bit gets flipped) = 0.2, hence P(particular bit is error free) = 0.8
Assuming error in each bit is independent of error in other bits,
P(0 bit gets flipped) = P(all bits are error free) = $(0.8)^7$
P(1 bit gets flipped) = $\binom{7}{1}*(0.2)*(0.8)^6$
Ans = 0.4233