In the arrangement depicted in the figure, a wire with a mass of...

For the air column in the tube, it is closed at one end (water surface) and open at the other. It vibrates in its fundamental mode.
The fundamental frequency of the closed organ pipe is given by:
$f_a = \dfrac{v_s}{4L_a}$
where $v_s = 340 \text{ m s}^{-1}$ is the speed of sound and $L_a = 1.0 \text{ m}$ is the length of the air column.
$f_a = \dfrac{340}{4 \times 1.0} = 85 \text{ Hz}$

The wire oscillates in its second harmonic and sets the air column into resonance, so the frequency of the wire is equal to the frequency of the air column.
$f_w = f_a = 85 \text{ Hz}$

The frequency of the $n$-th harmonic of a stretched string is:
$f_w = \dfrac{n}{2L_w} \sqrt{\dfrac{T}{\mu}}$
For the second harmonic, $n = 2$. The length of the wire is $L_w = 40 \text{ cm} = 0.4 \text{ m}$.
The linear mass density $\mu$ of the wire is:
$\mu = \dfrac{m}{L_w} = \dfrac{4.00 \times 10^{-3} \text{ kg}}{0.4 \text{ m}} = 10^{-2} \text{ kg m}^{-1}$

Substituting the values into the frequency formula:
$85 = \dfrac{2}{2 \times 0.4} \sqrt{\dfrac{T}{10^{-2}}}$
$85 = \dfrac{1}{0.4} \sqrt{100 T}$
$85 = 2.5 \times 10 \sqrt{T}$
$85 = 25 \sqrt{T}$
$\sqrt{T} = \dfrac{85}{25} = 3.4$

Squaring both sides to find the tension $T$:
$T = (3.4)^2 = 11.56 \text{ N}$

Rounding to one decimal place, we get $T \approx 11.6 \text{ N}$.