(b)

Conserving mechanical energy at the surface of earth and the maximum height attained,

$\frac{-G M m}{R_e}+\frac{1}{2} m \frac{4 G M}{3 R_e^2} R_e=\frac{-G M m}{r}+0$

$P_{. E_i}+K \cdot E_i=P_{.} E_f+K_i E_f$

$\Rightarrow \frac{-G M m}{R_e}+\frac{2 G M m}{3 R^e}=\frac{-G M m}{r}$

$-\frac{1}{3} \frac{G M m}{R_e}=\frac{-G M m}{r}$

$\Rightarrow r=3 R_e$

$\Rightarrow R_e+h=3 R_e$

$h=2 R_e$

Now, let us calculate the velocity of the particle at height equal to half of the maximum height $i . e$ at $h=R_e$ Again using mechanical conservation of energy,

$P . E_i+K . E_i=P . E_j+K . E_j$

$\frac{-G M m}{R_e}+\frac{1}{2} m \frac{4}{3} \frac{G M}{R_e^2} \times R_e=\frac{-G M m}{2 R}+\frac{1}{2} m v^2$

$\Rightarrow -\frac{1}{3} \frac{G M m}{R_e}+\frac{G M m}{2 R_e}=\frac{1}{2} m v^2$

$\Rightarrow \frac{G M m}{6 R_e}=\frac{1}{2} m v^2$

$\Rightarrow v=\sqrt{\frac{G M}{3 R_e}}=\sqrt{\frac{G M}{R_e^2} \times \frac{R_e}{3}}=\sqrt{\frac{g R_e}{3}}$