Answer
See the figure below.
Work Step by Step
Since the thermal energy increases, there is external work done here.
We know that the energy is conserved, so
$$E_i+W_{ext}=E_f$$
Thus,
$$K_i+U_i+W_{ext}=K_f+U_f+\Delta E_{th}$$
where the object loses 500 J as kinetic energy, gains 200 J as potential energy, and its thermal energy increases by 100 J.
$$500+0+W_{ext}=0+200+100$$
Thus, the external work is
$$W_{ext}=-200\;\rm J$$
See the figure below.
