Equation of wave, 

$y=0.2\sin [1.5x+60t]$

Comparing with standard equation, we get 

$y=A\sin [kx+\omega t]$

$k=1.5=\frac{2\pi }{\lambda }$

and $\omega =60=\frac{2\pi }{T}$

$\therefore$ Velocity of wave, $v=\frac{\omega }{k}$

$=\frac{60}{1.5}=40 \mathrm{m} {\mathrm{s}}^{-1}$

Velocity of wave in stretched string, $v=\sqrt{\frac{T}{m}}$

where, $m$ is the linear density, $T$ is tension in the string.

So, $T=v^{2}m=(40{)}^2\times 3\times {10}^{-4}$

$=0.48 \mathrm{N}$