Answer
The work done by the thrusters is $6.73\times 10^{8}~J$
Work Step by Step
We can find the change in mechanical energy when the lander moves to a higher orbit. Let $M_m$ be the moon's mass and let $M_l$ be the lander's mass. Let $R$ be the moon's radius.
$\Delta E_{mech} = \frac{1}{2}\Delta U$
$\Delta E_{mech} = \frac{1}{2}(U_f-U_0)$
$\Delta E_{mech} = \frac{1}{2}[-\frac{G~M_m~M_l}{R+300~km}-(-\frac{G~M_m~M_l}{R+50~km})]$
$\Delta E_{mech} = \frac{1}{2}(\frac{G~M_m~M_l}{R+50~km}-\frac{G~M_m~M_l}{R+300~km})$
$\Delta E_{mech} = \frac{1}{2}(G~M_m~M_l)(\frac{1}{R+50~km}-\frac{1}{R+300~km})$
$\Delta E_{mech} = \frac{1}{2}(6.67\times 10^{-11}~m^3/kg~s^2)(7.35\times 10^{22}~kg)(4000~kg)(\frac{1}{1.787\times 10^6~m}-\frac{1}{2.037\times 10^6~m})$
$\Delta E_{mech} = 6.73\times 10^{8}~J$
The work done by the thrusters is equal to the change in mechanical energy. Therefore, the work done by the thrusters is $6.73\times 10^{8}~J$
