Answer
$450\;\rm N$
Work Step by Step
The frequency of the wire is the same as the frequency of sound waves in the open-closed tube.
And since the wire vibrates at its fundamental frequency while the sound wave it generates excites the second vibrational mode,
$$f_{1,wire}=f_{3,sound}$$
where $m=3$ is the second vibrational mode for the open-closed tube.
$$\dfrac{v_{wire}}{2L_{wire}}=\dfrac{3v_{air}}{4L_{tube}}$$
Recalling that $v_{wire}=\sqrt{T_s/\mu}$, so
$$\dfrac{1}{ L_{wire}}\sqrt{\dfrac{T_s}{\mu}}=\dfrac{3v_{air}}{2L_{tube}}$$
$$ \sqrt{\dfrac{T_s}{\mu}}=\dfrac{3v_{air}L_{wire}}{2L_{tube}}$$
Multiplying both sides;
$$ \dfrac{T_s}{\mu}=\left[\dfrac{3v_{air}L_{wire}}{2L_{tube}}\right]^2$$
Thus,
$$ T_s=\mu\left[\dfrac{3v_{air}L_{wire}}{2L_{tube}}\right]^2$$
Plugging the known;
$$ T_s=(20\times 10^{-3})\left[\dfrac{3(340)(0.25) }{2(0.85)}\right]^2$$
$$T_s=\color{red}{\bf 450}\;\rm N$$
