Satellite Energy & Weightlessness question from MHT CET 2019 (2 May, Shift 2)

Answer: (b)

1. Energy just to raise it to altitude $2R$ ($r = 3R$): $\Delta U = GMm\left(\frac{1}{R} - \frac{1}{3R}\right) = \frac{2GMm}{3R}$.
2. If it must also orbit there, add $K = \frac{GMm}{6R}$, giving $\frac{5GMm}{6R}$.
3. Taking 'launch to an altitude $2R$' as only raising it, the answer is (b) $\frac{2GMm}{3R}$.