Answer
$4.8\;\rm cm$
Work Step by Step
This is a situation of a string that is fixed from both ends where the frequency is given by
$$f_m=\dfrac{mv}{2L}$$
And for the fundamental frequency
$$f_1=\dfrac{v}{2L}$$
Hence, for two different lengths of the string,
$$\dfrac{f_{1A}}{f_{1B}}=\dfrac{\frac{v}{2L_A}}{\frac{v}{2L_B}}=\dfrac{L_B}{L_A}$$
Noting that we assumed that the second given frequency is $f_{1B}$ when the length is $L_B$
Thus,
$$L_B=\dfrac{f_{1A}L_A}{f_{1B}}\tag 1$$
We need to find where must the violinist place his finger, which is the distance from its end
$$x=L_A-L_B$$
Plugging $L_B$ from (1);
$$x=L_A-\dfrac{f_{1A}L_A}{f_{1B}}=L_A\left[1-\dfrac{f_{1A} }{f_{1B}}\right]$$
Plugging the known;
$$x=30\left[1-\dfrac{440 }{523}\right]=4.76\;\rm cm$$
$$x=\color{red}{\bf 4.8}\;\rm cm$$
