1. Torsional frequency formula

For a body undergoing torsional oscillation, $f=\frac{1}{2\pi}\sqrt{\frac{C}{I}},$ where $C$ is the torsional constant of the wire and $I$ is the moment of inertia about the suspension axis.

So, if the torsional constant remains the same, then $f\propto \frac{1}{\sqrt{I}}.$

2. Initial moment of inertia of the rod

The rod has mass $M$ and length $2L$, suspended through its middle by a wire. The oscillation is about an axis through its centre and perpendicular to the rod.

Hence, $I_0=\frac{1}{12}M(2L)^2=\frac{1}{3}ML^2.$

3. Moment of inertia after attaching two masses

Two point masses, each of mass $m$, are attached symmetrically at distance $L/2$ from the centre.

Their total added moment of inertia is $I_{\text{added}}=2m\left(\frac{L}{2}\right)^2=2m\cdot \frac{L^2}{4}=\frac{1}{2}mL^2.$

Thus the new moment of inertia is $I_1=I_0+I_{\text{added}}=\frac{1}{3}ML^2+\frac{1}{2}mL^2.$

4. Use the 20% reduction in frequency

Frequency is reduced by $20\%$, so $f_1=0.8f_0.$

Since $f\propto 1/\sqrt{I}$, $\frac{f_1}{f_0}=\sqrt{\frac{I_0}{I_1}}=0.8.$

Squaring, $\frac{I_0}{I_1}=0.64,$ so $I_1=\frac{I_0}{0.64}=\frac{25}{16}I_0.$

Therefore, $I_1-I_0=\left(\frac{25}{16}-1\right)I_0=\frac{9}{16}I_0.$

But also, $I_1-I_0=\frac{1}{2}mL^2.$

Using $I_0=\frac{1}{3}ML^2$, $\frac{1}{2}mL^2=\frac{9}{16}\cdot \frac{1}{3}ML^2.$

Cancel $L^2$: $\frac{1}{2}m=\frac{3}{16}M.$

Hence, $m=\frac{3}{8}M.$

So, $\frac{m}{M}=\frac{3}{8}=0.375.$

5. Match with options

$\frac{m}{M}\approx 0.37$

So the correct option is C.