Answer
$70\space \mu m/s^{2}$
Work Step by Step
Please see the attached image first.
Let's apply the equation $V^{2}=U^{2}+2aS$ to the spacecraft to find it's acceleration as follows.
$\uparrow V^{2}=U^{2}+2aS$
Let's plug known values into this equation.
$0^{2}=(38\space cm/s)^{2}+2a\times(1\space km)$
Now convert the values into SI units using the conversion factors before solving the equation.
$0=(\frac{38\times10^{-2}\space m}{s})^{2}+ 2a\times 1000\space m$
$0=\frac{38^{2}\times10^{-4}\space m^{2}}{s^{2}}+2000a\space m$
$-1444\times10^{-4}\space m/s^{2}=2000a$
$a=\frac{-1444\times10^{-4}}{2000}\space m/s^{2}= -70\times10^{-6}\space m$
$a=-70\space \mu m/s^{2}$
Gravitational acceleration at the comet $= 70\space \mu m/s^{2}$
