This problem asks to to relate kinematics (motion) and dynamics (forces). This should instantly make you think Fnet=m*a. Mainily we are woried about the x component, though, so we'll say Fnet_x = m*a_x.

Now, you should know, that acceleration is a derivative of velocity, and that on a graph, the derivative is the slope. so a = dv/dx = rise/run = 6/2 = 3 m/s^2.

now, we have a mass of 3.0 kg, and an acceleration of 3m/s^2, so Fnet_x = 3*3 = 9N.

now, its a matter of adding the forces:

Fnet_x = F1_x + F2_x

F1 is along the X axis:

F1_x = F1 = 6.7N

F2 is not so we multiply its magnitude by cos(theta):

F2_x = F2*cos(th) = 9N*cos(th)

and we already know:

Fnet_x = 9N

we plug these back into the above equation:

9N = 6.7N + 9N*cos(theta)

now its an algebra problem:

9N*cos(th) = 9N - 6.7N = 2.3N

cos(theta) = 2.3N/9N = .255555555555

theta = arccos(.255555555555) = 75.19 degrees