= Solution
At each finite <stopping time> $\sigma_a^k$, the <Strong Markov property> says that $B_{\sigma_a^k+t}-B_{\sigma_a^k}$ is a standard <Brownian motion> independent of the past <sigma-algebra>. Its exit time from $(-a,a)$ is exactly $\sigma_a^{k+1}-\sigma_a^k$. By induction the <stopping times> are finite almost surely, and these durations are independent with common law $\sigma_a$, not merely identically distributed. This is the <Brownian exit-time skeleton>.
Put $v=\operatorname{Var}(\sigma_1)<\infty$, which follows from the moment estimate. <Brownian scaling> gives $\operatorname{Var}(\sigma_a)=a^4v$. For
$$
\tau_n=\sigma_{2^{-n}}^{2^{2n}},
$$
there are $2^{2n}$ independent durations of mean $C2^{-2n}$ and <variance> $v2^{-4n}$. Consequently
$$
\mathbb E\tau_n=C,\qquad\operatorname{Var}(\tau_n)=v2^{-2n}.
$$
For each $\varepsilon>0$, <Chebyshev's inequality> gives
$$
\sum_{n\geq0}\mathbb P(|\tau_n-C|>\varepsilon)
\leq\frac v{\varepsilon^2}\sum_{n\geq0}2^{-2n}<\infty.
$$
Apply the <Borel-Cantelli first lemma> for a countable sequence of positive errors tending to zero. We obtain the <Brownian exit-skeleton clock convergence>:
$$
\boxed{\sigma_{2^{-n}}^{2^{2n}}\longrightarrow C\quad\text{almost surely}.}
$$
No <independence> between the clocks for different values of $n$ is required.
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