Answer
$v_{fx} =\dfrac{\sqrt{ 2Fd m_1} }{ m_1+m_2 }$
Work Step by Step
Since the surface is frictionless, we can say that the system's [the two clay blobs] momentum is conserved.
The momentum of the system just before the first blob hits the second blob is given by
$$p_{ix}=p_{ix,1}+p_{ix,2}=m_1v_{1x}+m_{2}\overbrace{v_{ix,2}}^{=0}$$
$$p_{ix}= m_1v_{1x} \tag 1$$
And the final momentum of the system, after the first blob hits and sticks with the second blob and moves as one unit, is given by
$$p_{fx}=p_{fx,1}+p_{fx,2}=(m_1+m_2)v_{fx}$$
$$p_{fx}= (m_1+m_2)v_{fx}\tag 2$$
Since the momentum is conserved, so (1)= (2);
$$m_1v_{1x} =(m_1+m_2)v_{fx}\tag 3$$
Now we need to find the velocity of the first blob just before it hits the stationary one.
$$\sum F_x=F=m_{1}a_x$$
whereas $F$ is the constant pushing force.
Thus,
$$a_x=\dfrac{F}{m_1}\tag 4$$
Now we can use the kinematic formula of velocity squared.
$$v_{1x}^2=\overbrace{v_{ix}^2}^{=0}+2a_x\Delta x$$
where $\Delta x=d$;
$$v_{1x}^2= 2a_x d$$
Plugging from (4);
$$v_{1x}^2= \dfrac{2F}{m_1} d$$
Thus,
$$v_{1x} = \sqrt{ \dfrac{2Fd}{m_1} }$$
Plugging into (3);
$$m_1\sqrt{ \dfrac{2Fd}{m_1} }=(m_1+m_2)v_{fx} $$
Hence,
$$\boxed{v_{fx} =\dfrac{m_1}{(m_1+m_2)} \left(\sqrt{ \dfrac{2Fd}{m_1} }\right)}$$
$$v_{fx} =\dfrac{1}{(m_1+m_2)} \left(\sqrt{ \dfrac{2Fdm_1^{\color{red}{\bf\not} 2}}{\color{red}{\bf\not} m_1} }\right)$$
$$\boxed{v_{fx} =\dfrac{\sqrt{ 2Fd m_1} }{ m_1+m_2 }} $$
