Answer:
a) v =  1.19 m / s
, b)   P₁ = 0.922 10⁵ Pa
Explanation:
1) Let's use the fluid continuity equation
        Q = A v
The area of a circle is
       A = π r2 = π d²/4
       
      v = Q / A = Q 4 / pi d²
      v = 0.006 4/π 0.08²
      v =  1.19 m / s
2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point
      P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rho g y₂
The exercise tell us
P₂ = 1.0013 105 Pa
v₁ = 0
y₁ = 1 m
y₂=0  
Rho (water) = 1000 kg / m³
       P₁ + rho y₁ = P₂ + ½ rho v₂²
       P₁ = P₂ + ½ rho v₂² - rho g y₁
       P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1
       P₁ = 1.013 10⁵ +708.5  - 9800
       P₁ =  92208.5Pa
       P₁ = 0.922 10⁵ Pa