Answer
$r=0.57m$
Work Step by Step
Because the tension in the string provides the centripetal force $F_c$, $F_c$ cannot exceed $105N$. In other words, $$F_{c, max}=105N$$ $$m_{object}a_c=mr_f\omega_f^2=105N$$ $$r_f=\frac{105}{m\omega_f^2} (1)$$ This equation represents the radius of the smallest possible circle. We know $m$ but we do not know $\omega_f$ yet.
We take the initial point to be when the object moves in a circle of radius $r_0=1m$. 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), $$r_f=\frac{105}{\frac{mr_0^4\omega_0^2}{r_f^4}}=\frac{105r_f^4}{mr_0^4\omega_0^2}$$ $$r_f^3=\frac{mr_0^4\omega_0^2}{105}$$ $$r_f=\sqrt[3]{\frac{mr_0^4\omega_0^2}{105}}$$
We have $m=0.5kg$, so $r_0=1m$ and $\omega_0=6.28rad/s$
Therefore, $r_f=0.57m$
