Answer
$\rm0.411\; m/s$ at an angle of $47 ^\circ$ south of east.
Work Step by Step
The momentum is conserved just before and just after the collision of the two clay balls.
$$p_i=p_f$$
$$m_1v_{1i}+m_2v_{2i}=(m_1+m_2)v_f\tag 1$$
where the two balls stick together after the collision.
Now we need to write the initial velocities of the two clay balls in component form in which the positive $x$-direction is toward the east and the positive $y$-direction is toward the north.
$$v_{i1}=(2\;\hat i+0\;\hat j)\;\rm m/s\tag 2$$
$$v_{i2}=(-1.0\cos30^\circ \;\hat i-1.0\sin30^\circ\;\hat j)\;\rm m/s\tag 3 $$
Plug (2) and (3) into (1);
$$m_1(2\;\hat i+0\;\hat j)+m_2(-1.0\cos30^\circ \;\hat i-1.0\sin30^\circ\;\hat j)=(m_1+m_2)v_f $$
Plugging the known;
$$20(2\;\hat i+0\;\hat j)+30(-1.0\cos30^\circ \;\hat i-1.0\sin30^\circ\;\hat j)=(20+30)v_f $$
$$40 \;\hat i - 30 \cos30^\circ \;\hat i-30\sin30^\circ\;\hat j =50v_f $$
$$14.02\;\hat i -30\sin30^\circ\;\hat j =50v_f $$
Thus,
$$v_f=\dfrac{14.02\;\hat i -30\sin30^\circ\;\hat j }{50}$$
$$v_f=\left(\dfrac{14.02}{50}\right)\;\hat i -\left(\dfrac{30\sin30^\circ}{50}\right) \;\hat j\tag 4$$
So, the magnitude of the final velocity of the 50-gram ball is given by the Pythagorean theorem.
$$|v_f|=\sqrt{v_{fx}^2+v_{fy}^2}$$
Plugging from (4);
$$|v_f|=\sqrt{\left(\dfrac{14.02}{50}\right)^2+\left(\dfrac{-30\sin30^\circ}{50}\right)^2}$$
$$|v_f|=\color{red}{\bf 0.411}\;\rm m/s$$
And its direction is in the fourth quadrant since it has a positive $x$-direction component and a negative $y$-direction component.
And it is given by
$$\tan\theta=\dfrac{v_y}{v_x}$$
$$\theta=\tan^{-1}\left(\dfrac{v_y}{v_x}\right)$$
Plugging from (4);
$$\theta=\tan^{-1}\left(\dfrac{\left|\dfrac{-30\sin30^\circ}{50}\right|}{\dfrac{14.02}{50}}\right)\approx\color{red}{\bf 47^\circ}\tag{South of East}$$
So the 50-g clay ball will have a speed of 0.411 m/s at an angle of 47$^\circ$ south of east.
