A

Step 1: To find the maximum height reached by a projectile, we use the formula:
$$H = \frac{u^2 \sin^2 \theta}{2g}$$
where H is the maximum height, u is the initial velocity, \theta is the angle of projection, and g is the acceleration due to gravity.

Step 2: Let the initial velocity (u) be the same for both cases. Let the maximum height for the body projected at 30° be H1, and for the body at 60° be H2.

Step 3: Calculate H1:
$$H1 = \frac{u^2 \sin^2(30°)}{2g} = \frac{u^2 \left(\frac{1}{2}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{4}}{2g} = \frac{u^2}{8g}$$

Step 4: Calculate H2:
$$H2 = \frac{u^2 \sin^2(60°)}{2g} = \frac{u^2 \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{u^2 \cdot \frac{3}{4}}{2g} = \frac{3u^2}{8g}$$

Step 5: Now, find the ratio:
$$\frac{H1}{H2} = \frac{\frac{u^2}{8g}}{\frac{3u^2}{8g}} = \frac{1}{3}$$

Step 6: Therefore, the ratio of the maximum heights is: H1 : H2 = 1 : 3.
Thus, the answer is Option A.