= Solution
Each of the three input values has two possible outputs, so their union has at most six elements. The preceding union formula also gives this bound. Choose noise values $0,3$, whose two translated input <alphabets> are disjoint. With columns ordered $0,1,2,3,4,5$ the <transition matrix> is
$$
M=\begin{pmatrix}
1/2&0&0&1/2&0&0\\
0&1/2&0&0&1/2&0\\
0&0&1/2&0&0&1/2
\end{pmatrix}.
$$
Thus \b[$|\mathcal Y|=6$]. Every output determines its input, explicitly $X=Y\bmod3$, so $H(X\mid Y)=0$. Consequently $I(X;Y)=H(X)\leq\log_2 3$, with equality for a uniform input. Therefore
$$
\boxed{C=\log_2 3\ \text{bits per use},\qquad p_X=(1/3,1/3,1/3).}
$$
The extra uncertainty of which noise value occurred does not reduce information about the input: the two possible outputs for each input remain distinguishable from those of every other input.
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