Answer
The smallest distance from the axis is $0.17m$
Work Step by Step
Because static friction $f_s$ provides the centripetal force $F_c$, when $F_c\gt f_s^{max}$, the block will not stay in place anymore. $$F_{c}=f_s^{max}$$ $$m_{block}a_c=\mu_sF_N$$ $$mr_f\omega_f^2=\mu_smg$$ $$r_f\omega_f^2=\mu_sg$$ $$r_f=\frac{\mu_sg}{\omega_f^2} (1)$$ This equation represents the smallest distance from the axis the block can be relocated.
The initial point is when the block is resting at a distance $r_0=0.3m$ from the axis. Because angular momentum is conserved, $$I_0\omega_0=I_f\omega_f$$ $$mr_0^2\omega_0=mr_f^2\omega_f$$ $$r_0^2\omega_0=r_f^2\omega_f$$ $$\omega_f=\frac{r_0^2\omega_0}{r_f^2}$$
Plug it into (1), $$r_f=\frac{\mu_sg}{\frac{r_0^4\omega_0^2}{r_f^4}}=\frac{\mu_sgr_f^4}{r_0^4\omega_0^2}$$ $$r_f^3=\frac{r_0^4\omega_0^2}{\mu_sg}$$ $$r_f=\sqrt[3]{\frac{r_0^4\omega_0^2}{\mu_sg}}$$
We have $r_0=0.3m$, $\omega_0=2.2rad/s$, $\mu_s=0.75$ and $g=9.8m/s^2$
Therefore, $r_f=0.17m$
