Answer
$0.85\;\rm m/s$, $72.4^\circ$ south of east.
Work Step by Step
We know that the momentum is conserved before and after the collision.
So,
$$p_i=p_f$$
$$m_1v_{i1}+m_2v_{i2}+m_3v_{i3}=(m_1+m_2+m_3)v_f$$
We know that the three clay balls stick together after the collision.
where $m_1=40$ g, $m_2=30$ g, and $m_3=20$ g.
$$40v_{i1}+30 v_{i2}+20v_{i3}=(40+30+20)v_f$$
Thus,
$$4v_{i1}+3 v_{i2}+2v_{i3}=9v_f\tag 1$$
Now we need to write the initial velocities of the three clay balls in components form in which the positive $x$-direction is toward the east and the positive $y$-direction is toward the north.
$$v_{i1}=(4\cos45^\circ\;\hat i-4\sin45^\circ\;\hat j )\;\rm m/s$$
$$v_{i2}=(-3\;\hat i+0\;\hat j )\;\rm m/s$$
$$v_{i2}=(0\;\hat i+2\;\hat j )\;\rm m/s$$
Plugging these last 3 formulas into (1) and solving for $v_f$;
$$4(4\cos45^\circ\;\hat i-4\sin45^\circ\;\hat j )+3(-3\;\hat i+0\;\hat j )+2(0\;\hat i+2\;\hat j )=9 v_f$$
$$16\cos45^\circ\;\hat i-16\sin45^\circ\;\hat j -9\;\hat i +4\;\hat j =9 v_f$$
Hence,
$$9v_f=2.314\;\hat i-7.314\;\hat j$$
$$ v_f=\left(\dfrac{2.314}{9}\right)\;\hat i+\left(\dfrac{-7.314}{9}\right)\;\hat j$$
The speed of the resulting ball is given by
$$|v_f|=\sqrt{v_x^2+v_y^2}=\sqrt{\left(\dfrac{2.314}{9}\right)^2+\left(\dfrac{-7.314}{9}\right)^2}$$
$$|v_f|=\color{red}{\bf 0.85}\;\rm m/s$$
And its direction is in the fourth quadrant since it has a positive $x$-component and a negative $y$-component.
And it is given by
$$\theta=\tan^{-1}\left(\dfrac{|v_y|}{v_x}\right)=\tan^{-1}\left(\dfrac{\left|-7.314 \right|}{2.314}\right)$$
$$\theta=\color{red}{\bf72.4^\circ}\tag{South of East}$$
