Answer
The magnitude of the thrust needed is $2.94\times10^4N$
Work Step by Step
1) Find deceleration $a$ so that the craft's velocity reaches $0$ the instant it touches the ground
We have initial velocity $v_0=18m/s$, final velocity $v=0$ and the vertical distance $y=165m$. Deceleration $a$ can be found by $$v^2=v_0^2+2ay$$ $$a=\frac{v^2-v_0^2}{2y}=-0.98m/s^2$$
2) We already assume downward to be the positive direction, which is also the direction of the weight of the craft $mg_{lunar}$. Opposing this motion is the thrust $T$, which is in negative direction (upward).
According to Newton's 2nd Law: $$mg_{lunar}-T=ma$$ $$T=m(g_{lunar}-a)$$
We have $m_{craft}=1.14\times10^4kg$, $g_{lunar}=1.6m/s^2$ and $a=-0.98m/s^2$
$$T=1.14\times10^4(1.6+0.98)=2.94\times10^4N$$
