Answer
$18\;\rm cm$
Work Step by Step
We know that the force exerted by the spring on the string is equal to the tension in the spring.
So,
$$F_{sp}=T_s$$
$$T_s=kx\tag 1$$
where $k$ is the spring constant and $x$ is the stretched distance by the spring.
We can assume that the frequency in both cases [three and two antinodes] is the same since the string oscillates by the same oscillating magnetic field.
Hence,
$$f_2=f_3$$
$$\dfrac{2v_2}{2L}=\dfrac{3v_3}{2L}$$
Hence,
$$2v_2=3v_3 $$
$$2\sqrt{\dfrac{T_{s2}}{\mu}}=3\sqrt{\dfrac{T_{s3}}{\mu}} $$
Hence, bysquaring both sides.
$$4T_{s2}=9T_{s3}\tag2$$
In the first case, where we have 3 antinodes, the tension is then
$$T_{s3}=kx_i=0.08k$$
and for the second case, where we have 2 antinodes, the tension is then
$$T_{s2}=kx_f$$
Plug these into (2);
$$4kx_f=9(0.08k)$$
Hence,
$$x_f=\dfrac{9\times 0.08}{4}=0.18\;\rm m$$
$$x_f=\color{red}{\bf 18}\;\rm cm$$
