The speed of a projectile at highest point is equal to the component of velocity of projection along the horizontal direction. The angle given with the vertical is $60°$ therefore, the angle with the horizontal will be $30°$.

Therefore, 

$V=u\cos \theta$  $\Rightarrow V=10\cos 30°=10\times \frac{\sqrt{3}}{2}$

$\Rightarrow V=5\sqrt{3} \mathrm{m}{ \mathrm{s}}^{-1}$