I would also approximate the ice in the rink as a volume of a rectangle. V=L _W_ H According to this site:
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30 m by 60 m is the size of the official Olympic games rink with an average of 2.5 cm for the thickness of the ice.
And as 1m is 100cm, 60m is 6000cm and 30m is 3000cm
Using the volume formula we get $V= 4.5\times10^7 cm^3$ for the ice.
Now plugging this into the density equation to get mass we get,
$M = 4.125\times10^4 kg$