Answer:
 v₁ = 1,606 10⁴ m / s
Explanation:
For this exercise we must use Bernoulli's equation, let's use index 1 for the nozzle on the stairs and index 2 the pump on the street
               P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² +ρ g y₂
              
The pressure when the water comes out is the atmospheric pressure
            P₁ = P_atm
 The difference in height between the street and the nozzle on the stairs is
            y₂-y₁ = 15 m
Now let's use the continuity equation
              v₁ A₁ = v₂ A₁
The area of a circle is
              A = π r² = π (d/2)²
             v₁ π d₁²/ 4 = v₂ π d₂²/ 4
             v₂ = v₁ d₁² / d₂²
Let's replace
            P₂-P_atm + ½ ρ [ (v₁ d₁² / d₂²)²- v₁² ] + ρ g (y2-y1) = 0
            P₂- P_Atm + ρ g (y₂-y₁) = ½ ρ v₁² [1- (d₁/d₂)⁴]
            v₁² [1- (d₁/d₂)⁴] = (P₂-P_atm) ρ / 2 + g (y₂-y₁) / 2
Let's reduce the magnitudes to the SI system
            d₂ = 3.37 in (2.54 10⁻² m / 1 in) = 8.56 10⁻² m
            d₁ = 2.02 in = 5.13 10⁻² m
Let's calculate
             v₁² [1- (5.13 / 8.56) 4] = 449.538 10³ 10³/2  -9.8 15/2
             v₁² [0.8710] = 2.2477 10⁸ - 73.5 = 2.2477 10⁸
             v₁ = √ 2.2477 10⁸ /0.8710
              v₁ = 1,606 10⁴ m / s