A rod of mass M and length 2L is suspended at its middle by a wire....

Let $C$ be the torsional constant of the wire.

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{C}{\dfrac{M{\left(2L\right)}^{2}}{12}}} = \dfrac{1}{2\pi}\sqrt{\dfrac{3C}{M . {L}^{2}}}$

After masses are attached,

${f}^{′} = \dfrac{1}{2\pi}\sqrt{\dfrac{C}{\dfrac{M . {\left(2L\right)}^{2}}{12} + \dfrac{m{L}^{2}}{4} \times 2}}$

$⇒ 0.8f = \dfrac{1}{2\pi}\sqrt{\dfrac{C}{\left(\dfrac{M}{3} + \dfrac{m}{2}\right){L}^{2}}}$

$⇒ 0.64 \times \dfrac{3C}{M} = \dfrac{C}{\dfrac{M}{3} + \dfrac{m}{2}}$

\left(0.64\right)M + 0.64 \times \dfrac{3}{2}m = M$<br /><br />$⇒ \dfrac{m}{M} = 0.37