∴△h=△pρg= 100pa 1.0× 103kg/m3× 10n/kg= 1 100m，

The exposed volume of the plot is: vdew = s △ h =10-2m×1100m =10-4m3,

Boiling water volume of the block: V row =V-V dew = 2×10-4m3-10-4m3 =10-4m3,

According to the lever balance condition,

(m empty bucket +m sand) g×OA=(m g-F floating )× ob,

(1kg+m sand) g× OA = (1.6kg× g-1.0x103kg/m3× g×10-4m3) × ob,

And because ao: ob = 1: 2,

Rice and sand = 2 kg.

So the answer is: 2.