Loss of kinetic energy of the lead bullet$=$Heat generated to raise its temperature from $127 °\mathrm{C}$ to $327 °\mathrm{C}$.

$40 \%$ loss of the kinetic energy of the lead bullet$=$Heat gained by the lead bullet to raise its temperature.

$0.4\left(\frac{1}{2}m{v}^2\right)=m\times \left[S_{\mathrm{Pb}}\times ∆T+m\times L_{\mathrm{Pb}}\right]$

$0.4\times \frac{1}{2}m{v}^2=m{S}_{\mathrm{Pb}}\left(T_2-t_1\right)+m{L}_{\mathrm{Pb}}$

$0.2m{v}^2=m\times \left[125\times \left(327-127\right)+2.5\times {10}^4\right]$

$v^2=5\times \left[125\times 200+2.5\times {10}^4\right]$

$v^2=5\times \left[25000+25000\right]$

$v=500 \mathrm{m} {\mathrm{s}}^{-1}$