Let $u$ and $\theta$ be the velocity of projection and
angle of projection, respectively.
Given that, horizontal component of velocity,
$u_x=u \cos \theta=3 \mathrm{~m} / \mathrm{s}$
Equation of projectile motion is given by
$y=12 x-\frac{3}{4} x^2$ ...(i)
We know,
General equation of projectile motion,
$y=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}$ ...(ii)
Comparing Eqs. (i) and (ii), we get
$\tan \theta=12$
$\Rightarrow \quad \frac{\sin \theta}{\cos \theta}=12$
$\sin \theta=12 \cos \theta$
Multiplying on both side by $(u)$
$u \sin \theta=12(u \cos \theta)=12 \times 3$
i.e. $u \sin \theta=36 \mathrm{~m} / \mathrm{s}$
Now, using the expression of range,
$\begin{aligned} & R=\frac{u^2 \sin 2 \theta}{g} \\ & R=\frac{2 u^2 \sin \theta \cos \theta}{g}\end{aligned}$
$R=\frac{2(u \sin \theta)(u \cos \theta)}{g}$ [Using identity $\sin 2 \theta=2 \sin \theta \cos \theta$ ]
Substituting the values, we get
$\begin{aligned} R & =\frac{2 \times 36 \times 3}{10} \\ & =21.6 \mathrm{~m}\end{aligned}$