Answer
$\Delta S_L' = +710 J/K$
Work Step by Step
From (a), the change in entropy of the reservoir is $\Delta S_L = -710 J/K$
Note that they are in equilibrium state. This means the magnitude of change of entropy in the block is the same as the reservoir. The negative sign indicates that energy is being released from the block.
In the reservoir, however, the heat is absorbed, so the sign will be positive as indication of absorbed heat.
$\Delta S_L' = +710 J/K$
