As given that,
$r_p$ (radius of perihelion $)=2 R$
$r_a$ (radius of apnelion $)=6 R$
Hence,
$$
\begin{gathered}
r_a=a(1+e)=6 R \\
r_p=a(1-e)=2 R
\end{gathered}
$$
On dividing (i) and (ii), we get
$$
\left(\frac{1+e}{1-e}\right)=\frac{6}{2}
$$
eccentricity, $e=\frac{1}{2}$
There is not external force or torque on system.
So, by the law of conservation of angular momentum, angular momentum
at perigee $=$ angular momentum at apogee $L_1=L_2$
$\left(\because m_a=m_p=m\right.$, mass of satellite)
$\therefore m v_p r_p=m v_a r_a$
$\therefore \frac{v_a}{v_p}=\frac{1}{3}$ or $v_a=\left(\frac{v_p}{3}\right)$
where $m$ is mass of the satellite.
Applying conservation of energy,
energy at perigee $=$ energy at apogee
$$
\frac{1}{2} m v_p^2-\frac{G M m}{r_p}=\frac{1}{2} m v_a^2-\frac{G M m}{r_a}
$$
Multiplying $\frac{2}{m}$ to both sides,
$$
\begin{aligned}
&v_p^2-\frac{2 G M}{r_p}=v_a^2-\frac{2 G M}{r_a} \quad\left(\because v_a=\frac{v_p}{3}\right) \\
&v_p^2-\left(\frac{v_p}{3}\right)^2=-2 G M\left(\frac{1}{r_a}-\frac{1}{r_p}\right)
\end{aligned}
$$
where $M$ is the mass of the earth.
$$
\begin{aligned}
&\therefore v_p^2\left(1-\frac{1}{9}\right)=-2 G M\left(\frac{1}{r_a}-\frac{1}{r_p}\right) \\
&\Rightarrow v_p=\left[\frac{\frac{2 G M}{R}\left(\frac{1}{2}-\frac{1}{6}\right)}{\left(1-\frac{1}{9}\right)}\right]^{1 / 2} \\
&\left(\because r_p=2 R \text { and } r_a=6 R\right) \\
&=\left(\frac{2 / 3}{8 / 9} \frac{G M}{R}\right)^{1 / 2}=\sqrt{\frac{3}{4} \frac{G M}{R}} \\
&=\sqrt{\frac{3 \times 6.67 \times 10^{-11} \times 6 \times 10^{24}}{4 \times 6400 \times 10^3}}=6.85 \mathrm{~km} / \mathrm{s} \\
&
\end{aligned}
$$
$$
v_p=6.85 \mathrm{~km} / \mathrm{s}, v_a=\left(\frac{v_p}{3}\right)=2.28 \mathrm{~km} / \mathrm{s}
$$
For circular orbit of radius $r$,
$$
\begin{aligned}
&v_c=\text { orbital velocity }=\sqrt{\frac{G M}{r}} \\
&\text { For } r=6 R, \\
&v_c=\sqrt{\frac{G M}{6 R}} \sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6 \times 6.4 \times 10^6}} \\
&=3.23 \mathrm{~km} / \mathrm{s} .
\end{aligned}
$$
Hence, to transfer to a circular orbit at apogee, we have to boost the velocity by $\Delta v=(3.23-2.28)=0.95 \mathrm{~km} / \mathrm{s}$. This can be done by suitably firing rockets from the satellite.