A sphere of radius 1.00 cm is positioned in the path of a parallel...

The effective area of the sphere intercepting the parallel beam of light is its projected cross-sectional area, $A = \pi r^2$.

Given $r = 1.00 \text{ cm}$, the area is $A = \pi (1.00)^2 = \pi \text{ cm}^2$.

The total power absorbed by the sphere is $P = I A$.

Substituting $I = 0.50 \text{ W cm}^{-2}$, we get $P = 0.50 \times \pi = 0.50 \pi \text{ W}$.

Since the sphere completely absorbs the incident radiation, the force exerted by the light beam is the momentum transferred per unit time, $F = \dfrac{P}{c}$.

Using $c = 3 \times 10^8 \text{ m s}^{-1}$:

$F = \dfrac{0.50 \pi}{3 \times 10^8} = \dfrac{1.57}{3 \times 10^8} \approx 5.2 \times 10^{-9} \text{ N}$