How much would 125,000 solar panels generate electricity in a country were the daily mean temperaure is 18.6 and precipitation is 55.33 cm per...

This is a problem in scientific estimateion. Since the
problem isn't very specific we will have to use reasonable
numbers.

First we need to estimate how much power this
array might possibly generate.  Since there is no size given for the panels we will
estimate that each panel is 2 sq. meters in area.  This gives 2x12500=25,000 sq.meters
of collecting area.  Panels are typically 10%-15% efficicent, meaning that only a
fraction of the solar energy striking the panel is converted to electrical work.  We
will use the higher 15% estimate.  Solar energy flow (power) is 1000 kW per sq. meter. 
The best we can get out of this array is:

25,000 x 1000 x
.15=3.75x10^6 kW

We will use the usual electrical work unit
of kW-hr.  Now all we need to do is estimate how many hours per day the panels are in
operation over the course of a year.  55.33 cm per year is not a high number of inchs of
precipitation per year (roughly 20 inches/year) and 18.6 C average temperature would
place our panel in the drier midwest climates.  We will assume that on the 365-84=281
days of the year with no precipitation the panels are working all day.  The angle the
Sun strikes the panels is important as the power absorbed goes down with the sine of the
strike angle.  Also the latitude plays a part in that the Sun is not high overhead for
much of the day or the year.  We can estimate that all these factors together might
reduce the power output by 50% without further information.  This means that the total
hours of operation are 281x12x.5=1686 hours per year.  This gives a yearly estimate of
electrical work as:

1686x3.75x10^6=6.3x10^9
kW-hr

The per capita power consumption in the US in 2007
was 13,652 kw-hr per year.  This means that this array can supply yearly power
for

6.3x10^9/13652=463,000 households for an entire year. 
That's seems pretty impressive.