= Solution
For a nonempty <bounded set>, let $N_\delta(E)$ be the least number of radius-$\delta$ balls covering $E$. The lower and upper <Minkowski dimensions> are respectively
$$
\underline{\dim}_{\mathrm M}E=\liminf_{\delta\downarrow0}\frac{\log N_\delta(E)}{\log(1/\delta)},\qquad\overline{\dim}_{\mathrm M}E=\limsup_{\delta\downarrow0}\frac{\log N_\delta(E)}{\log(1/\delta)}.
$$
Using comparable cubes or ball diameters gives the same limits. For an unbounded set one must specify a local or bounded-piece convention; these are the usual bounded-set definitions used in the <Kakeya set> problem.
For the digit set, fixing the first $k$ digits gives exactly $3^k$ cylinders in intervals of length $9^{-k}$. They cover the set with at most $3^k$ balls of comparable radius. Their base-nine prefix integers are separated by at least two, so these containing intervals have gaps at least $9^{-k}$. Choosing one point in each cylinder shows that a radius-$9^{-k}$ ball meets at most a bounded number of chosen points. Hence $N_{9^{-k}}(Q)\asymp3^k$. For $9^{-(k+1)}<\delta\leq9^{-k}$, monotonicity sandwiches the <covering number> between constant multiples of $3^k$ and $3^{k+1}$. This proves the <box-counting dimension of separated digit sets> formula here:
$$
\boxed{\underline{\dim}_{\mathrm M}Q=\overline{\dim}_{\mathrm M}Q=\frac{\log3}{\log9}=\frac12.}
$$
A planar <Besicovitch set> is a <bounded set> of zero <Lebesgue measure> containing a unit line segment in every unoriented direction. The following proof in fact applies to every bounded planar <Kakeya set>, whether its area is zero or positive.
Choose $M\asymp\delta^{-1}$ directions equally spaced in an angular interval of length, say, $\pi/2$, and one segment of the set in each direction. Let $T_j$ be its rectangle of length one and width comparable to $\delta$, lying in the $C\delta$ neighborhood $E_{C\delta}$. Put $F=\sum_j1_{T_j}$. Then $\int F\asymp M\delta\asymp1$. Two such <Kakeya tubes> at angle $\alpha$ have overlap at most $C\delta^2/\alpha$, since the intersection of their supporting strips is a parallelogram of that area. The trivial bound $C\delta$ applies for nearly parallel tubes. Thus, uniformly over their translations,
$$
|T_j\cap T_k|\leq\frac{C\delta^2}{\delta+\alpha_{jk}}\lesssim\frac{\delta}{1+|j-k|}.
$$
Sum these intersections. Each row contributes at most $C\delta\log(2/\delta)$, so $\int F^2\lesssim\log(2/\delta)$. The <Cauchy-Schwarz inequality> on the union now gives the <planar Kakeya neighborhood lower bound>
$$
|E_{C\delta}|\geq\left|\bigcup T_j\right|\geq\frac{(\int F)^2}{\int F^2}\gtrsim\frac1{\log(2/\delta)}.
$$
A cover of $E$ by $N_\delta(E)$ radius-$\delta$ balls covers $E_{C\delta}$ by their fixed-factor enlargements. Hence $N_\delta(E)\gtrsim\delta^{-2}/\log(2/\delta)$. Boundedness gives the matching upper exponent $N_\delta(E)\lesssim_E\delta^{-2}$. Taking the lower and upper limits proves
$$
\boxed{\underline{\dim}_{\mathrm M}E=\overline{\dim}_{\mathrm M}E=2.}
$$
The logarithmic neighborhood loss is compatible with zero area; it still forces full dimension.
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