Answer
$ 2.52\times 10^5\;\rm m/s$
Work Step by Step
To solve this problem, we can use the principle of conservation of momentum.
$$p_i=p_f$$
Initially, the alpha particle has only kinetic energy in the $x$-direction while the $\rm ^{197}Au$ nucleus is at rest.
$$m_\alpha v_{ix,\alpha}+m_{\rm Au}v_{ix,\rm Au}=m_\alpha v_{fx,\alpha}+m_{\rm Au}v_{fx,\rm Au}$$
$$m_\alpha v_{i ,\alpha}+0=m_\alpha v_{f,\alpha}\cos49^\circ+m_{\rm Au}v_{f,\rm Au}\cos\theta$$
$$m_\alpha v_{i ,\alpha}-m_\alpha v_{f,\alpha}\cos49^\circ=m_{\rm Au}v_{f,\rm Au}\cos\theta\tag 1$$
For $y$-direction.
$$m_\alpha v_{iy,\alpha}+m_{\rm Au}v_{iy,\rm Au}=m_\alpha v_{fy,\alpha}-m_{\rm Au}v_{fy,\rm Au}$$
$$0+0=m_\alpha v_{f,\alpha}\sin49^\circ -m_{\rm Au}v_{f,\rm Au}\sin\theta$$
Hence,
$$v_{f,\rm Au}=\dfrac{ m_\alpha v_{f,\alpha}\sin49^\circ }{m_{\rm Au}\sin\theta}\tag 2$$
Plug into (1),
$$m_\alpha v_{i ,\alpha}-m_\alpha v_{f,\alpha}\cos49^\circ=m_{\rm Au}\dfrac{ m_\alpha v_{f,\alpha}\sin49^\circ }{m_{\rm Au}\sin\theta} \cos\theta$$
$$m_\alpha v_{i ,\alpha}-m_\alpha v_{f,\alpha}\cos49^\circ= \dfrac{ m_\alpha v_{f,\alpha}\sin49^\circ }{ \tan\theta} $$
Solving for $\theta$;
$$ \tan\theta= \dfrac{ m_\alpha v_{f,\alpha}\sin49^\circ }{ m_\alpha v_{i ,\alpha}-m_\alpha v_{f,\alpha}\cos49^\circ } $$
$$ \theta=\tan^{-1} \left[\dfrac{ v_{f,\alpha}\sin49^\circ }{ v_{i ,\alpha}- v_{f,\alpha}\cos49^\circ } \right]$$
Plug the known;
$$ \theta=\tan^{-1} \left[\dfrac{ (1.49\times 10^7)\sin49^\circ }{ (1.5\times 10^7)- (1.49\times 10^7)\cos49^\circ } \right]=\color{red}{\bf 65.1}^\circ $$
Plug into (2),
$$v_{f,\rm Au}=\dfrac{ m_\alpha v_{f,\alpha}\sin49^\circ }{m_{\rm Au}\sin65.1^\circ} $$
where $ m_\alpha=4u$, and $m_{\rm Au}=197u$
$$v_{f,\rm Au}=\dfrac{ 4 (1.49\times 10^7)\sin49^\circ }{197\sin65.1^\circ} $$
$$v_{f,\rm Au}=\color{red}{\bf 2.52\times 10^5}\;\rm m/s$$
