Answer:
a = 5.037 10⁻² m / s²
Step-by-step explanation:
For this exercise we will use the concept of radiation pressure for a reflective surface 
 
 P = S / c 
Where c is the speed of light and S is the pointing vector which is equal to the intensity of the radiation 
The definition of Pressure is 
 P = F / A 
 F = P A 
Newton's second law gives the relationship between force and acceleration 
 F = m a 
We substitute 
 P A = m a 
 a = I A / c m
 
Let's look for the radiation intensity, for this we use that the intensity in the Earth's orbit is 1360 W / m². 
 
We must look for the intensity in the orbit would want 0.5 AU, for this we use the definition of instance 
 I = P / A 
 
Where P is the power of the sun's emission that is constant 
 P = I A 
 I₁ A₁ = I₂ A₂
 
Where index 1 is for Earth and 2 for the orbit at 05 AU 
he area of a sphere is 
 A = 4 π R² 
 I₂ = I₁ A₁ / A₂ 
 I₂ = I₁ R₁² / R₂² 
 I₂ = 1360 (150 10⁶/75 10⁶)² 
 I₂ = 5440 W 
This is the radiation at the point of interest 
 
Let's calculate 
 a = 5440/3 10⁸ 10⁶/360 
 a = 5.037 10⁻² m / s²