Answer:
![intensity = 12.98 W/m^2](https://img.qammunity.org/2020/formulas/physics/college/xfxpff5n16rbklxbnqdeu5xyqu8cwdkieh.png)
![Energy = 3.93 J](https://img.qammunity.org/2020/formulas/physics/college/7d81z6yfew58rixogrwqt13ka9wjdhmkxv.png)
Step-by-step explanation:
As we know that the magnitude of electric field intensity is given as
![E = 98.9 V/m](https://img.qammunity.org/2020/formulas/physics/college/ilw8b066q288wqcx5nls81hqfpyoq02rsy.png)
now we know that intensity of the wave is given as the product of energy density and speed of the wave
![intensity = (1)/(2)\epsilon_0 E^2 c](https://img.qammunity.org/2020/formulas/physics/college/4zyt1zg63y112j2nes1njkgoqs1qya0b6u.png)
![intensity = (1)/(2)(8.85 * 10^(-12))(98.9)^2(3* 10^8)](https://img.qammunity.org/2020/formulas/physics/college/dezeoju0uazs6ynv480id7uhvwp65crkok.png)
![intensity = 12.98 W/m^2](https://img.qammunity.org/2020/formulas/physics/college/xfxpff5n16rbklxbnqdeu5xyqu8cwdkieh.png)
so intensity is the energy flow per unit area per unit of time
so the energy that flows through the area of 0.0259 m^2 in 11.7 s is given as
![Energy = Area * time * intensity](https://img.qammunity.org/2020/formulas/physics/college/e5xcqt337g08frvecle4mk3w1g89qpajp5.png)
![Energy = 0.0259(11.7)(12.98)](https://img.qammunity.org/2020/formulas/physics/college/56o6reicqr25fc3sze2f07t6wbxsf263yj.png)
![Energy = 3.93 J](https://img.qammunity.org/2020/formulas/physics/college/7d81z6yfew58rixogrwqt13ka9wjdhmkxv.png)