= Solution
Using the diameter normalization, define $\mathcal H^d_\delta(E)=\inf\{\sum_j(\operatorname{diam}U_j)^d:E\subseteq\bigcup_jU_j,\ \operatorname{diam}U_j\le\delta\}$ and the <Hausdorff measure> $\mathcal H^d(E)=\lim_{\delta\downarrow0}\mathcal H^d_\delta(E)$. The <Hausdorff dimension> is $\inf\{d:\mathcal H^d(E)=0\}$, equivalently $\sup\{d:\mathcal H^d(E)=\infty\}$.
The <Cantor set> is the intersection of the sets left by successively removing open middle thirds from $[0,1]$. At level $k$ it has $2^k$ <Cantor cylinders> of diameter $3^{-k}$. For $s=\log2/\log3$, their total $s$-cost is $2^k3^{-ks}=1$, so $\mathcal H^s(C)\le1$.
The random series $\xi=\sum_{n\ge1}X_n3^{-n}$ converges absolutely, and its ternary digits are all zero or two. Thus $\xi\in C$ and its law is the <Cantor Bernoulli measure>. Each level-$k$ cylinder has probability $2^{-k}$. Distinct such intervals have separation at least $3^{-k}$; a set $U$ of smaller diameter meets at most one. If $3^{-(k+1)}\le\operatorname{diam}U<3^{-k}$, then
$$
\mathbb P(\xi\in U)\le2^{-k}=2(3^{-(k+1)})^s\le2(\operatorname{diam}U)^s.
$$
Sets of diameter zero have probability zero, since the measure has no <measure atoms>; diameters at least one give the bound trivially. For arbitrary nonmeasurable sets use outer probability. Any countable cover of $C$ therefore satisfies $1\le\sum_j\mathbb P^*(\xi\in U_j)\le2\sum_j(\operatorname{diam}U_j)^s$. Hence
$$
\boxed{\tfrac12\le\mathcal H^s(C)\le1,\qquad\dim_H C=\frac{\log2}{\log3}.}
$$
The last implication follows by multiplying the $s$-costs by powers of the cover diameter: measures above $s$ vanish and those below $s$ are infinite.
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