I agree with B for this one.

Statement (1) gives us that a^2+b^2 is an integer, but that may or may not mean that the sqrt is an integer.

However, Statement (2) tells us that a^2 = 3(b^2). Then we can substitute this into the original equation

sqrt(a^2+b^2) = sqrt(3(b^2)+(b^2)) = sqrt(4(b^2)) = 2b

Since we are given that b is an integer, 2b will also be an integer. Hence, sqrt(a^2+b^2) = 2b is an integer.