Answer:
 g ±Δg = (9.8 ± 0.2) m / s²
Explanation:
For the calculation of the acceleration of gravity they indicate the equation of the simple pendulum to use
           T =  
           T² =   4pi2 L / g
4pi2 L / g
           g =  
They indicate the average time of 20 measurements 1,823 s, each with an oscillation
let's calculate the magnitude
            g =  4 pi2 0.823 / 1.823 2
4 pi2 0.823 / 1.823 2
             g = 9.7766 m / s²
now let's look for the uncertainty of gravity, as it was obtained from an equation we can use the following error propagation
for the period
              T = t / n
              ΔT =  Δt +
 Δt +  ΔDn
 ΔDn
In general, the number of oscillations is small, so we can assume that there are no errors, in this case the number of oscillations of n = 1, consequently
               ΔT = Δt / n
               ΔT = Δt
now let's look for the uncertainty of g
              Δg =  ΔL +
 ΔL +  ΔT
  ΔT
              Δg =  ΔL + 4π²L  (-2  T⁻³) ΔT
   ΔL + 4π²L  (-2  T⁻³) ΔT
             
a more manageable way is with the relative error
               
we substitute
               Δg = g ( \frac{\Delta L }{L} + \frac{1}{2}  \frac{\Delta T}{T}DL / L + ½ Dt / T)
the error in time give us the stanndard deviation  
let's calculate
                Δg = 9.7766 ( )
)
                Δg = 9.7766 (0.001215 + 0.0184)
                Δg = 0.19 m / s²
the absolute uncertainty must be true to a significant figure
                 Δg = 0.2 m / s2
therefore the correct result is
                g ±Δg = (9.8 ± 0.2) m / s²