If the satellite was orbiting the Earth in a circular orbit of radius $2a$, then its time period would be

$T=2\pi \sqrt{\frac{{\left(\mathrm{2}\mathit{a}\right)}^3}{GM}}$

The same would be the time period of the satellite if it orbits in an ellipse having a semi-major axis $2a$

The area of the ellipse is

$A=\pi \times 2a\times a=2\pi a^2$

$\frac{dA}{dt}=\frac{L}{2m}=\frac{2\pi a^2}{T}$, where $L$ is the angular momentum of the satellite about the centre of the Earth.

$\Rightarrow L=\frac{4\pi m{a}^2\sqrt{GM}}{2\pi \sqrt{8a^3}}=m\sqrt{\frac{GMa}{2}}$