The question is, which number does not have a radical sign after you simplify it.

So... the square root of #1# is #1#, so #sqrt(1)# is rational.

The square root of #2# cannot be simplified further, because #2# is not a perfect square. #sqrt(2)# is not rational.

#sqrt(65) = sqrt(5*13)#. This still has a radical sign and we cannot simplify it further, so this isn't rational.

#sqrt(196) = sqrt(4*49) = sqrt(2^2*7^2)=14#

#sqrt(196)# is rational, because we get a whole number without a radical#.^1#

#sqrt(225) = sqrt(25*9)=sqrt(5^2*3^2)=15#

#sqrt(225)# is rational, because we get a whole number without a radical.

So, the rational radicals are: #sqrt(1)#, #sqrt(196)# and #sqrt(225)#.

Footnote #1#: Not all rational numbers have to be whole ones. For example, #0.bar(11)# is rational, because it can simplify into a fraction. All rational numbers are by definition, a number that can simplify into a fraction. So, whole numbers are rational, but not all rational numbers are whole.