Answer
(a) On the system
(b) $-670J$
Work Step by Step
(a) We know that
$\Delta U=Q-W$
The internal energy increases, that is $\Delta U \gt 0$
$\implies Q-W\gt 0$
In an adiabatic process there is no exchange of heat, so the heat released is zero and thus $-W\gt 0$
$\implies W\lt 0$
This shows that the work is done "on the system".
(b) We know that
$\Delta U=Q-W$
We plug in the known values to obtain:
$670J=0-W$
$\implies W=-670J$
