The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed at 2m away from it, is :

The source emits light equally in all directions, so the energy spreads over the surface of a sphere of radius $$r$$.

Energy flux (intensity) at distance $$r$$ is given by the inverse-square law:
$$I = \frac{P}{4 \pi r^{2}}$$ where $$P$$ is the power of the source.

Substitute $$P = 450 \text{ W}$$ and $$r = 2 \text{ m}$$:
$$I = \frac{450}{4 \pi (2)^{2}} = \frac{450}{16 \pi}$$ W m$$^{-2}$$.

Numerical value:
$$I = \frac{450}{50.265} \approx 8.96 \text{ W m}^{-2}$$.

For a perfectly reflecting surface, radiation pressure is twice that for a perfectly absorbing surface. The formula is
$$P_{\text{rad}} = \frac{2I}{c}$$ where $$c = 3 \times 10^{8} \text{ m s}^{-1}$$.

Insert the value of $$I$$:
$$P_{\text{rad}} = \frac{2 \times 8.96}{3 \times 10^{8}}$$ Pa
$$= \frac{17.92}{3 \times 10^{8}} \text{ Pa}$$.

Simplify:
$$P_{\text{rad}} \approx 5.97 \times 10^{-8} \text{ Pa} \approx 6 \times 10^{-8} \text{ Pa}$$.

Therefore, the radiation pressure on the perfectly reflecting surface is $$6 \times 10^{-8}$$ Pascals.

Correct option: Option C.