Answer
$T=36N$
Work Step by Step
As it is clear from problem statement that there will be four loops, the length of the string is equal to four half wavelengths.
$L=4(\frac{\lambda}{2})=2\lambda$
$\lambda=\frac{L}{2}=\frac{0.90m}{2}=0.45m$
Frequency and wavelength are mutually related as follows
$v=f\lambda$
$v=(60)(0.45)=27\frac{m}{s}$
Linear mass density of string can be determined as follows
$\mu=\frac{0.044Kg}{0.90m}=4.9\times10^{-2}\frac{Kg}{m}$
We also know that
$v^2=\frac{T}{\mu}$
or $T=v^2\mu$
putting the values, we get
$T=(27)^2(4.9\times10^{-2})$
$T=36N$
Thus tension in string is $36N$
