The braking tension force for the wire is, $F=\sigma ·A$

$$\begin{gathered} \Rightarrow F=\left(4.8\times {10}^7 \mathrm{N} {\mathrm{m}}^{-2}\right)\left({10}^{-6} {\mathrm{m}}^2\right) \\ \Rightarrow F=48 \mathrm{N} \end{gathered}$$

Now, the tension force will provide the necessary centripetal force so that it can rotate in a horizontal circle,

$$\begin{gathered} F_C=48 \mathrm{N} \\ m{\omega }^{2}r=48 \mathrm{N} \\ \omega =\sqrt{\frac{48}{10\times 0.3}}=4 \mathrm{rad} {\mathrm{s}}^{-1} \end{gathered}$$

Thus, the maximum angular velocity with which it can rotate in a horizontal circle is $4 \mathrm{rad} {\mathrm{s}}^{-1}$