A sector of 56°, cut out from a circle, contains 17.6 cm2.

Area of the sector =17.6 cm2

Area of the sector $=\frac{\pi r^{2} \theta}{360}$

$\Rightarrow 17.6=\frac{22}{7} \times r^{2} \times \frac{56}{360}$

$\Rightarrow r^{2}=\frac{17.6 \times 7 \times 360}{22 \times 56}$

$\Rightarrow r^{2}=36$

$\Rightarrow r=6 \mathrm{~cm}$

∴ Radius of the circle = 6 cm