If mass of the bullet is $\mathrm{m}\, \mathrm{gm}$.

 then total heat required for bullet to just melt clown

$\mathrm{Q}_{1}=\mathrm{m} \mathrm{c} \Delta \theta+\mathrm{mL}=\mathrm{m} \times 0.03(327-27)+\mathrm{m} \times 6$

$=15 \mathrm{m} \mathrm{cal}=(15 \mathrm{m} \times 4.2) \mathrm{J}$

Now when bullet is stopped by the obstacle, the loss in its mechanical energy $=\frac{1}{2}\left(\mathrm{m} \times 10^{-3}\right) \mathrm{v}^{2} \mathrm{J}$

(As $\left.\mathrm{mg}=\mathrm{m} \times 10^{-3} \mathrm{kg}\right)$

As $25 \%$ of this energy is absorbed by the obstacle. The energy absorbed by the bullet

$\mathrm{Q}_{2}=\frac{75}{100} \times \frac{1}{2} \mathrm{mv}^{2} \times 10^{-3}=\frac{3}{8} \mathrm{mv}^{2} \times 10^{-3} \mathrm{J}$

Now the bullet will melt if $\mathrm{Q}_{2} \geq \mathrm{Q}_{1}$

i.e., $\frac{3}{8} \mathrm{mv}^{2} \times 10^{-3} \geq 15 \mathrm{m} \times 4.2 \Rightarrow \mathrm{v}_{\min }=410 \mathrm{m} / \mathrm{s}$