The way to do it using your reasoning is as follows:
After t seconds speed is 9.8 x t. So average speed over that interval is 9.8xt/2 as you say in your question (if t=1 sec then av. speed is 9.8/2, and if t=2 then av. speed =9.8 etc).
So now the average speed after travelling for t seconds is distance/time.
So after it has travelled 1 metre its average speed is 1/t.
But above we have shown that average speed is also equal to 9.8 x t/2.
So this means that 1/t=9.8 x t/2 after 1 metre.
Rearranging this gives t^2=2/9.8, so t=srt(2/9.8)=approx 0.45
The thing to realise is that the average speed over the first second is not the same as the average speed over the second second:
Av. speed over first second is 9.8/2 metres/sec as you say.
Since after 2 seconds the speed will be 9.8 x 2 then the average speed over the 2nd second is 1/2(9.8 x 2+9.8)=3 x 9.8/2 metres/sec. So the average speed of the ball over the 2nd second is 3 times the average speed over the first second.
That's why your calculation doesn't work, because you are assuming that the average speed is the same over the same period of time, and it isn't.