The initial speed of a projectile fired from ground is $$u$$. At the highest point during its motion, the speed of projectile is $$\dfrac{\sqrt{3}}{2}u$$. The time of flight of the projectile is:

At the highest point of projectile motion, the vertical component of velocity is zero, so the speed equals the horizontal component:

$$ v_{\text{top}} = u\cos\theta $$

Given that $$v_{\text{top}} = \dfrac{\sqrt{3}}{2}u$$:

$$ u\cos\theta = \dfrac{\sqrt{3}}{2}u $$

$$ \cos\theta = \dfrac{\sqrt{3}}{2} $$

$$ \theta = 30° $$

The time of flight of a projectile is:

$$ T = \dfrac{2u\sin\theta}{g} = \dfrac{2u\sin 30°}{g} = \dfrac{2u \times \frac{1}{2}}{g} = \dfrac{u}{g} $$