A block of mass 1.9 kg rests on a smooth horizontal table and is attached to a fixed wall by a light spring of force constant 400 N m^-1 along its…

Collision is instantaneous and the spring force is finite, so linear momentum of the bullet-block system is conserved during the impact: $0.1\times200=(1.9+0.1)v$, giving $v=10\ \mathrm{m\,s^{-1}}$. The block is at the natural length at that instant, which is the equilibrium position, so the speed there is the maximum speed of the oscillation. Mechanical energy is conserved afterwards: $\frac12(2.0)v^2=\frac12kA^2$, so $A=v\sqrt{m/k}=10\sqrt{2/400}=10\times0.07071=0.7071\ \mathrm{m}$. In centimetres this is $70.71$. Kinetic energy is not conserved in the impact (it falls from $2000$ J to $100$ J), so equating the bullet's energy to the spring energy would give a wrong amplitude.