Answer
$5.4\;\rm m$
Work Step by Step
Since the string vibrates at its third-harmonic frequency, the wavelength of the standing waves is then equal to two-thirds the length of the string; $(\lambda=\frac{2}{3}L)$.
The amplitude is given by
$$A_{(x)}=A_{max}\sin (kx)$$
where $k=2\pi/\lambda$
$$A_{(x)}=A_{max}\sin \left(\dfrac{2\pi}{\frac{2}{3}L}x\right)$$
$$A_{(x)}=A_{max}\sin \left(\dfrac{3\pi}{L}x\right)$$
At $x=30$ cm, $A=\frac{1}{2}A_{max}$, so
$$\frac{1}{2}\color{red}{\bf\not} A_{max} =\color{red}{\bf\not} A_{max}\sin \left(\dfrac{3\pi}{L}[0.30]\right)$$
$$ \dfrac{0.9\pi}{L} =\sin^{-1}\left[\frac{1}{2}\right]$$
$$ \dfrac{0.9\color{red}{\bf\not} \pi}{L} = \frac{\color{red}{\bf\not} \pi}{6} $$
Thus,
$$L=\color{red}{\bf 5.4}\;\rm m$$
