No. The heater will only replace exactly the amount of heat lost(Unless it's too small to maintain the temperature)
If that's an accurate average then you'd be using an average of 100 watts of power or 2.4 kw/hr per day. Roughly $.50/day.
Energy required to heat 200L of water to 25c is straightforward math assuming a starting temp of 15C and assuming no losses of heat during the process. Heater wattage is irrelevant.
It takes 4.2 joules to heat 1 gram of water 1 degree c.
So you're heating 200,000 grams 10 degrees. So, 200,000*10*4.2=8,400,000 joules.
One joule is 1 watt second or 3,600,000 joules per kilowatt hour. So 8,400,000/3,600,000= 2.3 kw hours to heat 200L costing $.40-$.50
Only effect in a perfect situation the wattage of the heater would have is the length of time it would take, in this case is that it would take roughly 7.6 hours.
Calculating mathematically the heat loss of the tank though is more complex. You'd need the outside air temp, glass thickness, glass area, adjustments for heat lost through the base, to air through the lid, etc.
Might be easier to get a watt meter throw it on the heater for a week. (Something like $12 at the warehouse) Then you'll know the power usage and can calculate the average rate of loss. Watts of electricity used=watts of heat lost (Resistive heaters are 100% efficient)