I think you mean that your cylindrical tank is lying on its side on the ground. If this is the case then the work done to let the water run out on the ground will be negative and will be numerically equal to the gravitational potential energy of the water in the tank.

First imagine that the tank is half burried. In this state it has (on average) nil gravitational potential and a mass of kg

Now imagine lifting the tank until it sits fully on the ground. This is a 1 m lift.

GPE added=mgh=10000pi*9.81*1

In SI units the density of water is 1 kg per litre or 1000 kg per m^3.