Answer
$2.35\times 10^{-4}\;\rm N$
Work Step by Step
We know, for standing waves in a string, that the frequency is given by
$$f_m=\dfrac{mv}{2L}$$
And the fundamental frequency $(m=1)$ is then given by
$$f_1=\dfrac{v}{2L}$$
where $v=\sqrt{T/\mu}$ where $T$ is the tension and $\mu=m/L$ is the linear density of the string.
Thus,
$$f_1=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}\tag 1$$
Note that the density of the string is $ρ=mV=m/AL$ where $m$ is the mass of the string, $L$ is the length of the string, and $A$ is its cross-sectional area. And we know that $m/L=\mu$
Thus,
$$\mu=\rho A$$
Plugging into (1):
$$f_1=\dfrac{1}{2L}\sqrt{\dfrac{T}{\rho A}} $$
Solving for $T$;
$$2f_1L=\sqrt{\dfrac{T}{\rho A}} $$
$$4f_1^2L^2=\dfrac{T}{\rho A} $$
Hence,
$$T=4f_1^2L^2\rho A=4f_1^2L^2\rho( \pi r^2)$$
$$T=4\pi f_1^2L^2 r^2\rho$$
Plugging the known;
$$T=4\pi (100)^2(0.12)^2 (10\times 10^{-6})^2(1300)$$
$$T=\color{red}{\bf 2.35\times 10^{-4}}\;\rm N$$
