Explanation:
The given data is as follows.
  Thickness = 0.8 cm = [tex]0.8 \times 10^{-3} m[/tex]   (As 1 m = 1000 cm)
  Diameter = 14.0 cm,  Conductivity = [tex]220 W/(m^{o}C)[/tex]
  mass = 1.2 g,    L = [tex]2.26 \times 10^{6} J/kg[/tex]
Now, we will calculate the heat of vaporization for water as follows.
        Q = mL
      Q = [tex]0.8 \times 10^{-3} \times 2.26 \times 10^{6}[/tex]
      Q = 1808 J
So, rate of heat transfer in 1 sec  will be as follows.
    [tex]\frac{Q}{t}[/tex] = 1808 J/s
Thus, we can conclude that 1808 J/s heat is transfered to the boiling water in one second.