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Surface area problem

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Question:

Assuming that 68.6% of the earth's surface is covered with water at an average depth of 1.01 miles, estimate the mass of the water on earth in kg.

Given the following:

Density of water = 1000 kg/m^3
1 mile = 1.6093 km
1.01 miles = 1.6254 km

Total surface area of earth = 509,600,000 km^2. Area of water = 509,600,00 * 68.6% = 349,585,600 km^2. The volume of water is 349,585,600 km^2 * 1.6254 km = 568,213,987 km^3. Divide by 1000 to get m^3 = 568213.987m^3.

Then, 568213.987m^3 * 1000kg/m^3 = 5.68213987E8 kg. This was wrong by my guess. Can you help me figure out what went wrong?

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Question:

Assuming that 68.6% of the earth's surface is covered with water at an average depth of 1.01 miles, estimate the mass of the water on earth in ...

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