A piano wire A vibrates with a fundamental frequency of 600 Hz . A...

The initial fundamental frequency of wire A and the identical wire B is $f_B = 600 \text{ Hz}$.

When the tension in wire A is slightly increased, its frequency increases. Since it produces $6$ beats per second with wire B, the new frequency of wire A is:

$f_A = 600 + 6 = 606 \text{ Hz}$

The fundamental frequency of a stretched string is given by $f = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$, which implies $f \propto \sqrt{T}$.

Therefore, the ratio of the tension in A to the tension in B is:

$\dfrac{T_A}{T_B} = \left(\dfrac{f_A}{f_B}\right)^2$

$\dfrac{T_A}{T_B} = \left(\dfrac{606}{600}\right)^2 = (1.01)^2 = 1.0201 \approx 1.02$