
We have R=πL $\begin{aligned}
& \mathrm{g}0=\frac{2 \mathrm{G} \frac{\mathrm{M}}{\mathrm{L}}}{\mathrm{R}}=\frac{2 \mathrm{GM} \pi}{\mathrm{L}^2} \
& \therefore \mathrm{F}{\mathrm{m}}=\mathrm{mg}_0=\frac{2 \mathrm{GM} \pi \mathrm{m}}{\mathrm{L}^2}
\end{aligned}$
Hence option (4)