A string of length $$L$$ is fixed at one end and carries a mass of $$M$$ at the other end. The mass makes $$\left(\frac{3}{\pi}\right)$$ rotations per second about the vertical axis passing through end of the string as shown. The tension in the string is $$\underline{\hspace{2cm}} ML.$$

Given frequency of rotation 

$$f=\frac{3}{\pi\ }$$

Now calculate angular velocity : -

$$\omega\ =2\pi\ f$$

$$\omega\ =2\pi\ \times\ \frac{3}{\pi\ }$$

$$\omega\ =2\times\ 3\ =\ 6$$

We know that centrifugal force will be equal to the horizontal component of the T(tension of the wire)

$$\frac{mv^2}{r}=T\sin\theta\ $$

$$v=\omega\ r$$

so 

$$m\omega^2r=T\sin\theta\ $$

Here $$r=L\sin\theta\ $$

$$m\omega^2L\sin\theta\ =T\sin\theta\ $$

$$m\omega^2L\ =T$$

substitute values

$$M\left(6\right)^2L\ =T$$

$$T=36ML\ $$

so answer is 36