I wanted to provide some details on the proof, please let me know any further details are required.
To prove that $M_t$ is a local martingale consider the following stopping times: $T_k=\inf\{t: R_t=\frac 1k\}$, $S_k=\inf\{t: R_t=k\}$, $\tau_k=T_k\wedge S_k\wedge n$. We know that $R_t$ satisfies $R_t=\int_{0^t}\frac{d-1}{2R_s}ds + B_t$, where $B_t$ is a Brownian Motion (see Karatzas, Shreve, proposition III.3.21). We can note that stopped processes $B_{t\wedge \tau_k}$ take values within $[1/k,k]$ where function $f$ is twice continuously differentable, thus allowing to apply Ito formula for $f(R_t)$. Doing this, because of the choice of the function $f(x)=x^{2-d}$, we see that $f(R_{t\wedge \tau_k})$ will be a local martingale satisfying
$$
M_{t\wedge \tau_k}=M_{1\wedge \tau_k}+(2-d)\int_{1}^{t\wedge \tau_k}R_s^{1-d}dB_s
$$
Using the fact that $\tau_k$ is bounded stopping time (because Bessel process of dimension $d\ge 3$ almost surely doesn't reach the origin and $R_t\to \infty$ $(t\to \infty)$) and boundedness of $R_s$ in $s\in (1,\tau_k)$ we conclude that $\int_{1}^{t\wedge \tau_k}R_s^{1-d}dB_s$ is a martingale. Using these arguments we also have that $\tau_k\uparrow \infty$.
Next, we can show that in dimension $d=3$, $M_t=R_t^{-1}$ is not a martingale by computing $\mathbb{E}(M_t)$.
$$
\mathbb{E}(M_t)=\int_{\mathbb{R}^3}\frac{1}{\sqrt{x^2+y^2+z^2}}\frac{1}{(\sqrt{2\pi t})^3}\exp\left(-\frac{x^2+y^2+z^2}{2t}\right)dxdydz
$$
Rewriting this integral in spherical coordinates $(x,y,z)\mapsto (r, \phi, \theta)$, we can rewrite it as
$$
\frac{1}{(\sqrt{2\pi t})^3}\int_{0}^\infty\int_{0}^{2\pi}\int_{0}^{\pi}\frac 1r\exp(-\frac{r^2}{2t})r^2\sin\theta drd\phi d\theta=
$$
$$
=\frac{1}{(\sqrt{2\pi t})^3}\int_{0}^\infty\frac 1r\exp(-\frac{r^2}{2t})r^2 dr\cdot 2\pi \int_{0}^{\pi}\sin \theta d\theta=
$$
$$
=\frac{2\cdot 2\pi t}{(\sqrt{2\pi t})^3}\int_{0}^\infty\exp(-u) du =\sqrt{\frac{2}{\pi t}}
$$
If $M_t$ were martingale then $\mathbb{E}(M_t)$ had to be the same for all $t\ge 1$, but above computation show that this expectation depends on $t$.
