A sector with acute central angle $$\theta$$ is cut from a circle of diameter 12 cm. The radius of the circle circumscribed about the sector is

ABC is sector that is cut from a circle of diameter 12 cm. 

So, AB = AC = $$\frac{12}{2} = 6 cm$$

Let AO = R = Radius of circle that circumscribed the sector. 

OM is perpendicular to AM. AM = $$\frac{AB}{2} = \frac{6}{2} = 3 cm$$

Since, $$\angle{BAC} = \theta, then \angle{BAO} = \frac{\theta}{2}$$

Now, 

In $$\triangle{AMO}:$$

$$sec\frac{\theta}{2} = \frac{AO}{AM}$$

$$sec\frac{\theta}{2} = \frac{R}{3}$$

$$R = 3sec\frac{\theta}{2}$$