Answer
$265\;\rm N$
Work Step by Step
First, we need to recognize between the two waves we got here, the one in the air from the one in the string.
We know that the tension in the string is given by
$$v_{string}=\sqrt{\dfrac{T_s}{\mu}}$$
Hence,
$$T_s =\mu v_{string}^2 \tag 1$$
We have here a standing wave from the violin string since it is fixed from both ends.
The frequency of the sound wave emitted from the violin string is the same frequency as the violin string wave.
The frequency of the sound wave emitted from the string is given by
$$f=\dfrac{v_{air}}{\lambda_{air}}\tag 2$$
And the frequency of the string wave is given by
$$f_m=\dfrac{mv_{string}}{2L}$$
And the fundamental frequency $(m=1)$ is then given by
$$f_1=\dfrac{v_{string}}{2L}\tag 3$$
And since $f_1=f$, and from (2) and (3),
$$\dfrac{v_{string}}{2L}=\dfrac{v_{air}}{\lambda_{air}} $$
Thus,
$$v_{string}=\dfrac{2Lv_{air}}{\lambda_{air}}$$
Plugging into (1);
$$T_s =\mu \left[\dfrac{2Lv_{air}}{\lambda_{air}}\right]^2 $$
Plugging the known;
$$T_s =(1\times 10^{-3})\left[\dfrac{2(0.30)(343) }{(0.40)}\right]^2 $$
$$T_s=\color{red}{\bf 265}\;\rm N$$
