= Solution
Let $\xi_j=B_{\sigma_1^j}-B_{\sigma_1^{j-1}}$. Continuity makes each displacement equal to $+1$ or $-1$. Reflection symmetry of a fresh <Brownian motion> gives <probability> one half to each sign, and the <Strong Markov property> makes these signs independent across successive exits. Thus $B_{\sigma_1^n}=\sum_{j=1}^n\xi_j$.
For the normalized position, the <characteristic function> is
$$
\mathbb E\exp\left(iu\frac{B_{\sigma_1^n}}{\sqrt n}\right)
=\left[\cos(u/\sqrt n)\right]^n\longrightarrow e^{-u^2/2},
$$
since $\log\cos z=-z^2/2+O(z^4)$ near zero. By the <Lévy continuity theorem>, this proves
$$
\boxed{B_{\sigma_1^n}/\sqrt n\ \Rightarrow\ N(0,1).}
$$
This is the <Gaussian limit of a Brownian exit skeleton>.
Compare along the subsequence $n=2^{2m}$ with $\tau_m=\sigma_{2^{-m}}^{2^{2m}}$. Joint <Brownian scaling> of the successive exits gives
$$
B_{\tau_m}\ \stackrel d=\ 2^{-m}B_{\sigma_1^{2^{2m}}}.
$$
The right-hand side converges weakly to $N(0,1)$. The preceding part and path continuity give $B_{\tau_m}\to B_C$ almost surely on the original Brownian <probability> space. Hence its weak limit is also the law $N(0,C)$. Uniqueness of the weak limit forces \b[$C=1$], so finally \b[$\mathbb E\sigma_a=a^2$]. <Independence> of the position and its random clock was not assumed in this comparison.
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