The entropy of a sample of a certain substance increases by $$0.836\ \text{J K}^{-1}$$ on adding reversibly $$0.3344\ \text{J}$$ of heat at constant temperature. The temperature of the sample is:

For a reversible isothermal process the entropy change is given by the fundamental relation
$$\Delta S = \frac{q_{\text{rev}}}{T}$$

The problem supplies
$$\Delta S = 0.836\ \text{J K}^{-1}, \quad q_{\text{rev}} = 0.3344\ \text{J}$$

Re-arranging the formula to find the temperature,
$$T = \frac{q_{\text{rev}}}{\Delta S} = \frac{0.3344}{0.836}\ \text{K}$$

Evaluating the ratio,
$$T = 0.4\ \text{K}$$

Hence, the temperature of the sample is 0.4 K.
Option D which is: 0.4 K