Correct Answer - B

Given, m = 500 kg

Rise in temperature, `dT=(100^(@)-20^(@)C = 80^(@)`C

Specific heat of water, c`=10^(3)` cal `kg^(-1) C^(-1)`

Heat energy gained by water, E=mcdT

`=500 xx 10^(3) xx 80`cal = `4 xx 10^(7) xx 4.2`J = `16.8 xx 10^(7)`J

From , `E=mc^(2)`

Increase in mass of water,

`m=E/c^(2) = (16.8 xx 10^(7))/(3 xx 10^(8))^(2) = 1.87 xx 10^(-9)` kg