Answer
$(d)$ the entropy of the warmer water decreases by the same amount that the entropy of the colder water increases.
Work Step by Step
The entropy equation is $\Delta S = \frac{Q}{T} $,
In Carnot engine, change of entropy is zero. So $\Delta S = 0$
This results in the ratio of heat in cold reservoir to the hot reservoir $Q_C/Q_H $ is always equals to the ratio of temperature of cold water to hot water $T_C/T_H$.
The relationship becomes $|\frac{Q_C}{Q_H}| = |\frac{T_C}{T_H}|$
and if we rearrange it becomes $|\frac{Q_C}{T_C}| = |\frac{Q_H}{T_H}|$
Hence $\Delta S_{Hot} = \Delta S_{Cold} $
This means the entropy of the warmer water decreases by the same amount that the entropy of the colder water increases. so choice $(d)$ is correct.
