Answer
The launch angle is $75.5^\circ$.
Work Step by Step
At the highest point. the vertical component of the projectile's velocity is equal to zero, while the horizontal component is the same as the initial. Because the vertical component is equal to zero at the highest point, $v_0/4$ is exactly equal to the horizontal component. Now, in the initial moment we have
$$\cos\theta=\frac{v_x}{v_0}=\frac{v_0/4}{v_0}=\frac{1}{4}$$
so the launch angle is
$$\theta=\arccos\frac{1}{4}=75.5^\circ.$$
