Answer:
 v_{f} = 115.95 m / s
Explanation:
This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions
         Thrust =  
          -v₀ = v_{e}
-v₀ = v_{e} 
where v_{e} is the velocity of the gases relative to the rocket
let's apply these expressions to our case
the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units
        M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg
      
The final mass is the mass of the engines + the mass of the rocket
       M_{f} = 25.5 +54.5 = 80 g = 0.080 kg
thrust and duration of ignition are given
        thrust = 5.26 N
        t = 1.90 s
Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear
           thrust = v_{e}  
           v_{e} = thrust   
           v_{e} = 5.26  
           v_{e} = - 786.93 m / s
the negative sign indicates that the direction of the gases is opposite to the direction of the rocket
now we look for the final speed of the rocket, which as part of rest its initial speed is zero
             v_{f}-0 = v_{e} 
we calculate
             v_{f} = 786.93 ln (0.0927 / 0.080)
             v_{f} = 115.95 m / s