I assume that calculators are not allowed, perhaps not even paper and pencil.
With paper and pencil, there is an algorithm to find the square-root of any number accurate to as many digits as we wish. Please post if that is the case.
Without calculator, pen and pencil, we resort to mental arithmetic. Someone with a little practice can memorize or calculate the square of all numbers up to 100, or as many as one can.
Here we can take advantage of exactly the property suggested by Xisca... that is, sqrt(44)=2sqrt(11), which reduces the calculation to more accuracy with a lesser knowledge of the squares of numbers.
As a first approximation, we know that 33*33=1089, or 3.3*3.3=10.89, which is already very close to 11.
Thus a first approximation to sqrt(44)=2sqrt(11)=2*3.3=6.6 (approximately).
To refine the calculation of sqrt(11), we use Newton's approximation, namely
a better approximation is given by 3.3 + (11 - 3.3^2) / (2*3.3) = 3.3+0.11/6.6 = 3.3+1/60
which gives
sqrt(44) = 2(3.3+1/60) = 6.6+1/30 = 6.6333,
actually accurate almost to the nearest 1/10000.
All of the above can be done mentally
For those who need help with rounding to the nearest hundredth (0.01), we drop digits AFTER the 2nd decimal, namely *.**33. If the first digit dropped is 5 or more, add one to the last digit retained, otherwise, retain only the two digits after the decimal. In this case, the first digit dropped is 3, so it will simply be dropped. The final answer is then 6.63, to the nearest hundredth.
denote the event that UTC beats Marshall, so that 
, and let 
denote the event that UTC beats Furman, so that 
. We're told that the UTC has a 
probability of beating both Marshall and Furman.