= Solution
Write $r=\|y\|$ and use the ordinary Euclidean <Laplacian> $\sum_i\partial_i^2$, so the generator of standard <Brownian motion> is half that operator. For a radial function $f(r)$, the <Laplacian in polar coordinates> is $f''(r)+(d-1)f'(r)/r$. With $f(r)=r^{2-d}$,
$$
f'(r)=(2-d)r^{1-d},\qquad f''(r)=(2-d)(1-d)r^{-d},\qquad\Delta h=0\quad(r>0).
$$
Thus $h$ is a <harmonic function> on the punctured space.
We first justify that the singularity is not visited. For $0<\eta<x<R$, stop at $S=\tau_\eta\wedge\tau_R$. The <Itô formula> makes $h(B_{t\wedge S})$ a <local martingale>, and it is bounded on this annulus, hence a true <martingale>. The exit time is finite almost surely: $\mathbb P(S>t)\leq\mathbb P(\|B_t\|<R)\to0$ by the Gaussian transition density. Taking the bounded terminal limit yields
$$
\mathbb P(\tau_\eta<\tau_R)\,\eta^{2-d}\leq h(\bar x)=x^{2-d}.
$$
If the origin were reached before $\tau_R$, every smaller radius would first be reached, so letting $\eta\downarrow0$ shows that event has probability zero. Taking a countable sequence $R\to\infty$ proves $\tau_0=\infty$ almost surely. Localization on increasing annuli now applies the <Itô formula> globally and gives the <Bessel power local martingale>
$$
h(B_t)=h(\bar x)+\int_0^t\nabla h(B_s)\cdot dB_s.
$$
Stopping this <continuous local martingale> at any $\tau_a$, including $a=0$, preserves its local-<martingale> property.
For $0<a<x$, the stopped path stays at radius at least $a$, so $0<M_t\leq a^{2-d}$. The <bounded local martingale criterion> makes it a true <martingale>. For $a=x$, Brownian radial oscillations at its starting sphere give $\tau_x=0$ almost surely and the stopped process is constant. Thus the same true-<martingale> conclusion holds for $0<a\leq x$.
For $a=0$, the process is a <strict local martingale>. To prove this, write $B_t=\bar x+\sqrt t\,G$ in distribution, with $G$ a standard $d$-dimensional normal vector. For every shift $v$,
$$
\mathbb E\|G+v\|^{2-d}\leq1+(2\pi)^{-d/2}\int_{\|u\|\leq1}\|u\|^{2-d}\,du=:C_d<\infty.
$$
Indeed the integrand outside the unit ball is at most one, while inside it the Gaussian density is at most $(2\pi)^{-d/2}$ and the radial integral is a constant times $\int_0^1r\,dr$. <Brownian scaling> therefore gives
$$
\mathbb E h(B_t)\leq C_dt^{-(d-2)/2}\longrightarrow0.
$$
A true <martingale> would have expectation $h(\bar x)>0$ at every time, which is impossible.
The printed range also permits $a>x$, and this case must not be grouped with the bounded case. Here $\tau_a<\infty$ almost surely, by the same Gaussian bound for exit from a bounded ball. Moreover,
$$
\mathbb E M_t=a^{2-d}\mathbb P(\tau_a\leq t)+\mathbb E\bigl[h(B_t)\mathbf1_{\{\tau_a>t\}}\bigr]\longrightarrow a^{2-d}<x^{2-d},
$$
since the second expectation is at most $\mathbb E h(B_t)\to0$. It too is a <strict local martingale>. The complete <stopped inverse radial power martingale classification> is therefore
$$
\boxed{M\text{ is a true martingale for }0<a\leq x;\quad M\text{ is strict local for }a=0\text{ or }a>x.}
$$
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