Total mass of ice in box `=4kg`
surface area of cubical box
`A=6(side)^(2) = 6 xx 30 xx 30 xx 10^(4) = 54 xx 10^(-2) m^(2)`
Thichness `d = 5 xx 10^(-2)`
Time `t = 6 xx 3600 =21,600 sec`
Let m is of ice melted in `6hr`
`(KA(Deltatheta)t)/(d) =mL_(ice)`
`(0.01 xx 54 xx 10^(-2) xx 45 xx 6 xx 3600)/(5 xx 10^(2)) = m xx 335 xx 10^(3)`
`rArr m =0.3 kg`
`:.` Mass of ice remaining after 6hrs
`=4kg - 0.3kg = 3.7Kg` .