Answer
$28.4\;\rm cm$
Work Step by Step
We have here a standing wave from the violin string since it is fixed from ts 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
$$f=\dfrac{v_{air}}{\lambda}\tag 1$$
And the frequency of the string wave is given by
$$f_m=\dfrac{mv_{string}}{2L}$$
where $v_{string}=\sqrt{T/\mu}$ where $T$ is the tension force in the string and $\mu$ is the linear density of the string.
So for the fundamental frequency;
$$f_1=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}\tag 2$$
And since $f_1=f$, and from (1) and (2),
$$\dfrac{v_{air}}{\lambda}=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}} $$
Solving for $L$;
$$L =\dfrac{\lambda}{2v_{air} }\sqrt{\dfrac{T}{\mu}} $$
Plugging the known;
$$L =\dfrac{(39.1\times 10^{-2})}{2(344) }\sqrt{\dfrac{(150)}{(0.600\times 10^{-3})}} $$
$$L=0.284\;\rm m=\color{red}{\bf 28.4}\;\rm cm$$
