Modulus of rigidity of a horizontal rod with hanging mass is given as
$$
\eta=\frac{F_L}{A x}=\frac{m g L}{A x}
$$

Where, $x=$ verticle deflection of the end of the rod.




Here, $L=6 \mathrm{~cm}=6 \times 10^{-2} \mathrm{~m}$, radius $R=2 \mathrm{~cm}$
$$
=2 \times 10^{-2} \mathrm{~m}
$$
mass, $m=400 \pi \mathrm{kg}$ and $\eta=3 \times 10^{10} \mathrm{~N} / \mathrm{m}^2$
Putting these values, in Eq. (i) we get
$$
\begin{aligned}
& 3 \times 10^{10}=\frac{(400 \pi \times 10) \times 6 \times 10^{-2}}{\pi\left(2 \times 10^{-2}\right)^2 x} \\
& \Rightarrow x=\frac{4 \pi \times 10^3 \times 6 \times 10^{-2}}{\pi \times 4 \times 10^{-4} \times 3 \times 10^{10}} \Rightarrow x=0.02 \mathrm{~mm}
\end{aligned}
$$
Hence, the correct option is (2).