Given data

• Intensity of sunlight: $I = 50\ \text{W/m}^2$
• Area: $A = 1\ \text{m}^2$
• Reflected part: $25\% = 0.25$
• Absorbed part: $75\% = 0.75$
• Speed of light: $c = 3 \times 10^8\ \text{m/s}$

Radiation pressure idea Force on the surface comes from change in momentum of light.

• For the absorbed part, pressure is: $P_{\text{abs}} = \frac{I_{\text{abs}}}{c}$
• For the reflected part, momentum change doubles, so pressure is: $P_{\text{ref}} = \frac{2I_{\text{ref}}}{c}$

Split the intensity $I_{\text{ref}} = 0.25 \times 50 = 12.5\ \text{W/m}^2$ $I_{\text{abs}} = 0.75 \times 50 = 37.5\ \text{W/m}^2$

Total pressure $P = \frac{I_{\text{abs}}}{c} + \frac{2I_{\text{ref}}}{c}$ $P = \frac{37.5}{3\times 10^8} + \frac{2(12.5)}{3\times 10^8}$ $P = \frac{37.5 + 25}{3\times 10^8}$ $P = \frac{62.5}{3\times 10^8}$

Force on $1\ \text{m}^2$ area Since $F = PA$ and $A=1$, $F = P = \frac{62.5}{3\times 10^8}$ $F \approx 2.08 \times 10^{-7}\ \text{N}$

Compare with options $2.08 \times 10^{-7}\ \text{N} = 20.8 \times 10^{-8}\ \text{N}$ Closest option is: A: $20 \times 10^{-8}\ \text{N}$

Final answer $\boxed{A}$