The expression for the maximum height of the projectile is given
by,

$H=\frac{u^2 {\sin }^2 \theta }{2 \mathrm{g}} \dots \left(1\right)$

The expression for the maximum range of the projectile is given by,

$R=\frac{u^2 \sin 2\theta }{g}$

The ratio of the maximum height to the range of the projectile is calculated as,

$\frac{H}{R}=\frac{\frac{u^2 {\sin }^2\theta }{2 \mathrm{g}}}{\frac{u^2 \sin 2\theta }{g}}$

$=\frac{{\sin }^3 \theta }{2 \sin 2\theta }$

Substitute the given values in the above equation.

$\frac{H}{R}=\frac{{\sin }^3 \theta }{2 \sin 2\theta }$

$\frac{3}{4}=\frac{{\sin }^3 \theta }{2 \sin 2\theta }$

$3 \sin 2\theta -2 {\sin }^2 \theta$

$\tan \theta =3$

From the geometry of the triangle,

$\sin \theta =\frac{3}{\sqrt{10}} ...\left(2\right)$

From the equation $\left(1\right)$

$\frac{u^2 {\sin }^2 \theta }{2 \mathrm{g}}=3$

$u^2 {\sin }^2 \theta =3\times 2 \mathrm{g}$

Substitute the values in the above equation.

$u^2 {\left(\frac{3}{\sqrt{10}}\right)}^2=3\times 2 \left(10\right)$

$u^2=\frac{100\times 2}{3}$

$u=10\sqrt{\frac{2}{3}} \mathrm{m} {\mathrm{s}}^{-1}$