Answer:
Q = 165.95 cm³ / s,  1)    v =  ,  2)  v = 2.05 m / s,
,  2)  v = 2.05 m / s,
 3)  d₂ = 1.014 cm
Explanation:
This is a fluid mechanics exercise
1) the continuity equation is
          Q = v A
 
where Q is the flow rate, A is area and v is the velocity
          
the area of a circle is
         A = π r²
radius and diameter are related
         r = d / 2
substituting
        A = π d²/4
        Q = π/4   v d²
let's reduce the magnitudes
        v = 0.55 m / s = 55 cm / s
let's calculate
        Q = π/4   55   1.96²
        Q = 165.95 cm³ / s
If we focus on a water particle and apply the zimematics equations
         v² = v₀² + 2 g y
where the initial velocity is v₀ = 0.55 m / s
         v =  
         v =  
2) ask to calculate the velocity for y = 0.2 m
         v =  
         v = 2.05 m / s
3) We write the continuous equation for this point 2
         Q = v₂ A₂
         A₂ = Q / v₂
let us reduce to the same units of the SI system
         Q = 165.95 cm³ s (1 m / 10² cm) ³ = 165.95 10⁻⁶ m³ / s
         A₂ = 165.95 10⁻⁶ / 2.05
         A₂ = 80,759 10⁻⁶ m²
area is
         A₂ = π/4   d₂²
         d₂ =  
         d₂ =  
         d₂ = 10.14 10⁻³ m
         d₂ = 1.014 cm