1. Given data

• Mass of body: $m = 10\,\text{kg}$
• Length of wire: $r = 0.3\,\text{m}$
• Breaking stress of wire: $\sigma = 4.8 \times 10^7\,\text{N m}^{-2}$
• Area of cross-section: $A = 10^{-2}\,\text{cm}^2$

2. Convert area into SI units

Since, $1\,\text{cm}^2 = 10^{-4}\,\text{m}^2$

Therefore, $A = 10^{-2} \times 10^{-4} = 10^{-6}\,\text{m}^2$

3. Maximum tension in the wire

Breaking stress is $\sigma = \frac{T_{\max}}{A}$

So, $T_{\max} = \sigma A = (4.8 \times 10^7)(10^{-6}) = 48\,\text{N}$

4. Condition for circular motion in space station

Since the body is rotated in a space station, effectively there is no gravity affecting the rotation. Hence the wire tension alone provides the centripetal force:

$T = m\omega^2 r$

At maximum angular speed, tension reaches breaking value: $T_{\max} = m\omega_{\max}^2 r$

Thus, $48 = 10\cdot \omega^2 \cdot 0.3$

$48 = 3\omega^2$

$\omega^2 = 16$

$\omega = 4\,\text{rad s}^{-1}$

5. Final answer

The maximum angular speed is: $\boxed{4}$