Answer
$\approx 500\;\rm m/s$
Work Step by Step
We can solve this problem by using the conservation of energy and momentum laws.
The energy is conserved after the collision when the bullet and the block are moving as one unit.
so
$$\color{red}{\bf\not} \frac{1}{2}kA^2=\color{red}{\bf\not} \frac{1}{2}(m_{\rm bullet}+m_{\rm block})v_1^2$$
where $v_1$ is their velocity after the collision.
$$v_1 = \sqrt{ \dfrac{kA^2}{m_{\rm bullet}+m_{\rm block}} }\tag 1$$
The momentum is conserved after and before the collision,
$$m_{\rm bullet}v_0=(m_{\rm bullet}+m_{\rm block})v_1$$
So,
$$v_0=\dfrac{m_{\rm bullet}+m_{\rm block}}{m_{\rm bullet}}v_1$$
Plugging from (1);
$$v_0=\dfrac{m_{\rm bullet}+m_{\rm block}}{m_{\rm bullet}}\sqrt{ \dfrac{kA^2}{m_{\rm bullet}+m_{\rm block}} } $$
$$v_0=\sqrt{\dfrac{(m_{\rm bullet}+m_{\rm block})^2}{m_{\rm bullet}^2} \dfrac{kA^2}{m_{\rm bullet}+m_{\rm block}} } $$
$$v_0=\dfrac{A}{m_{\rm bullet}}\sqrt{k(m_{\rm bullet}+m_{\rm block}) } $$
Plugging the known;
$$v_0=\dfrac{0.1}{(10\times 10^{-3})}\sqrt{2500([10\times 10^{-3}]+[1.0]) } $$
$$v_0=\color{red}{\bf 502}\;\rm m/s$$
