Answer
$4.1\times 10^7\;\rm m/s$
Work Step by Step
Using the conservation of energy principle and assuming that there are no external forces exerted on this system [the Alpha particle+ the thorium nucleus].
$$U_i+K_i=U_f+K_f$$
where $_i\rightarrow$ when the particle is still on the thorium nucleus surface, and $_f\rightarrow$ when it is far away from it.
The thorium nucleus remains constant in both cases, but the alpha particle was initially at rest, and finally when $r\rightarrow\infty$, its speed is $v_\alpha$.
$$\dfrac{kq_{\alpha}q_{\rm thorium}}{r_{\rm thorium}}+0=0+\frac{1}{2}m_{\alpha}v_{\alpha}^2$$
$$v_\alpha=\sqrt{\dfrac{2kq_{\alpha}q_{\rm thorium}}{m_{\alpha}r_{\rm thorium}}}$$
where $q_{\alpha}=2e$, and $q_{\rm thorium}=90e$
$$v_\alpha=\sqrt{\dfrac{360ke^2}{m_{\alpha}r_{\rm thorium}}}$$
Plug the known;
$$v_\alpha=\sqrt{\dfrac{360(9\times 10^9)(1.6\times 10^{-19})^2}{(4\times 1.67\times 10^{-27})(7.5\times 10^{-15})}}$$
$$v_\alpha=\color{red}{\bf 4.1\times 10^7}\;\rm m/s$$
