The area bounded by the curve $x^2=ky,k>0$, and the line $y=3$ is $12{\text{ units}}^2$. Then $k$ is:

The curve $x^2=ky$ is a parabola symmetric about the y-axis. The region is between $y=0$ and $y=3$.For a given $y$, $x=\pm \sqrt{ky}$.Area $A=2{\int }_0^{3}xdy$ (due to symmetry).
$$12=2{\int }_0^3\sqrt{k}{y}^{1/2}dy$$
$$6=\sqrt{k}{\left[\frac{y^{3/2}}{3/2}\right]}_0^3$$
$$6=\sqrt{k}{\left[\frac{2}{3}{y}^{3/2}\right]}_0^3$$
$$6=\frac{2\sqrt{k}}{3}(3\sqrt{3})$$
$$6=2\sqrt{k}\sqrt{3}$$
$$3=\sqrt{3k}$$
Squaring both sides:
$$9=3k⟹k=3$$