We are given an equation of stationary wave,

$Y=0.3\sin \left(0.157x\right)\cos \left(200\pi t\right)$.

Comparing it with the general equation of stationary wave,

 $Y=a\sin \left(kx\right)\cos \left(\omega t\right)$.

We get,

$k=\left(\frac{2\pi }{\lambda }\right)=0.157$

$\Rightarrow \lambda =\frac{2\pi }{0.157}=4{\pi }^2 \left(\because \frac{1}{2\pi }\approx 0.157\right)$.

As the possible wavelength associated with the $n^{\mathrm{th}}$ harmonics of a vibrating string fixed at both ends is given as,

$\lambda =\frac{2l}{n} \Rightarrow l=n\left(\frac{\lambda }{2}\right)$.

Now, according to the question, string is vibrating in the $4^{\mathrm{th}}$ harmonic, so,

$4\left(\frac{\lambda }{2}\right)=l\Rightarrow 2\lambda =l$

$\Rightarrow l=2\times 4{\pi }^2=8{\pi }^2$ [using equation $\left(1\right)$]

Now,

$\therefore {\pi }^2\approx 10$

$\Rightarrow l\approx 80 \mathrm{m}$.