Answer
$\approx 8.8\;\rm m/s$
Work Step by Step
We chose east to be the positive $x$-direction and north to be the positive $y$-direction and the impact point is the origin. This means that the final velocity is making an angle of 35$^\circ$ with the positive $x$-direction (counterclockwise).
Now we need to write the velocity of the cars into the components form.
The Cadillac car;
$$v_{iC}=(0\;\hat i+3\;\hat j)\;\rm m/s\tag 1$$
The Volkswagen car;
$$v_{iV}=(v_{ix,V}\;\hat i+0\;\hat j)\;\rm m/s\tag 2$$
The momentum is conserved just before and after the collision.
Thus,
$$p_i=p_f$$
Hence,
$$m_Cv_{iC}+m_Vv_{iV}=(m_C +m_V)v_f$$
Plugging from (1) and (2);
$$m_C(0\;\hat i+3\;\hat j)+m_V(v_{ix,V}\;\hat i+0\;\hat j)=(m_C +m_V)(v_{f}\cos\theta\; \hat i+v_{f}\sin\theta\;\hat j)$$
Thus,
$$3m_C\;\hat j +v_{ix,V} m_V \;\hat i =(m_C +m_V)(v_{f}\cos\theta\; \hat i+v_{f}\sin\theta\;\hat j)$$
Plugging the known;
$$3\times 2000\;\hat j +1000v_{ix,V} \;\hat i =(2000+1000)(v_{f}\cos\theta\; \hat i+v_{f}\sin\theta\;\hat j)$$
$$6000\;\hat j +1000v_{ix,V} \;\hat i =3000 v_{f}\cos\theta\; \hat i+3000 v_{f}\sin\theta\;\hat j $$
For $x$-direction;
$$1000v_{ix,V} \;\hat i =3000 v_{f}\cos\theta\; \hat i$$
$$ v_{ix,V} =3 v_{f}\cos35^\circ\tag 3$$
For $y$-direction;
$$6000\;\hat j =3000 v_{f}\sin\theta\;\hat j $$
$$2 = v_{f}\sin35^\circ $$
Thus,
$$v_f=\dfrac{2}{\sin35^\circ } $$
Plugging into (3);
$$ v_{ix,V} =3 \cos35^\circ\dfrac{2}{\sin35^\circ } =\color{red}{\bf 8.57}\;\rm m/s $$
Therefore, the initial speed of the Volkswagen car was 8.6 m/s eastward.
