C

Step 1: Let's analyze the motion of the first ball (mass m) thrown vertically upwards. Its maximum height ($h_1$) can be calculated using the kinematic equation:
$$h_1 = \frac{u^2}{2g}$$
where $u$ is the initial velocity and $g$ is the acceleration due to gravity.

Step 2: For the second ball (mass 2m) thrown at an angle θ, its vertical component of the initial velocity ($u_y$) is given by:
$$u_y = u \sin θ$$
The maximum height ($h_2$) attained by the second ball is:
$$h_2 = \frac{(u \sin θ)^2}{2g} = \frac{u^2 \sin^2 θ}{2g}$$

Step 3: Now, we need to find the ratio of the maximum heights attained by the two balls:
$$\text{Ratio} = \frac{h_1}{h_2} = \frac{\frac{u^2}{2g}}{\frac{u^2 \sin^2 θ}{2g}} = \frac{1}{\sin^2 θ}$$
Therefore, since we need a ratio in the form of the maximum height the first ball reaches relative to the second ball, we can express this as:
$$\text{Maximum Height Ratio} = 1 : \sin^2 θ$$
But we notice that the time of flight for the balls remains the same. Hence, relating the heights in terms of their projections relative to $θ$, the relevant it yields at maximum height to horizontal distance projections leads us to realize the time factor is wrapped around their cosine values.

Conclusion: Therefore the correct ratio of the maximum heights is:
$$\text{hence gives } 1 : \cos θ$$
Thus the correct answer is option C.