= Solution
The ideal observer performs <maximum a posteriori decoding>: for received word $y$, it maximizes $\Pr(X=c\mid Y=y)$, equivalently $\Pr(X=c)\Pr(Y=y\mid X=c)$. <Maximum-likelihood decoding> maximizes the latter conditional likelihood alone. <Minimum-distance decoding> minimizes the <Hamming distance> $d(c,y)$. Ties may be resolved arbitrarily among maximizers.
For independent errors in a <binary symmetric channel>,
$$
\Pr(Y=y\mid X=c)=p^{d(c,y)}(1-p)^{n-d(c,y)}.
$$
For $0<p<1/2$, its ratio on increasing the distance by one is $p/(1-p)<1$. Thus <maximum-likelihood decoding> and <minimum-distance decoding> select exactly the same codewords. At the allowed endpoint $p=0$, they agree on a possible observation $y\in C$; impossible observations have zero likelihood for every codeword, so an arbitrary maximum-likelihood tie rule need not agree with minimum distance. This qualification is needed for the literal assumption $p<1/2$.
For the specified observation, the <Hamming distances> from 000 and 111 are two and one. Their likelihoods are $3/64$ and $9/64$, respectively. After multiplying by the priors, their joint probabilities are $12/320$ and $9/320$, so their posterior probabilities are $4/7$ and $3/7$. Therefore
$$
\boxed{\text{ideal observer: }000,\qquad\text{maximum likelihood: }111,\qquad\text{minimum distance: }111}.
$$
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