For any two phases, the difference in their entropies at a temperature $$T$$ is obtained from their heat-capacity difference:

$$S_\beta - S_\alpha = \displaystyle\int_{0}^{T}\frac{C_{P,\beta}-C_{P,\alpha}}{T}\,dT = \int_{0}^{T}\frac{\Delta C_P}{T}\,dT$$

Given that $$\Delta C_P=C_{P,\beta}-C_{P,\alpha}=1\ \text{J mol}^{-1}\,\text{K}^{-1}$$ is constant (200-700 K), we first find the enthalpy difference between the phases.

Enthalpy difference up to any temperature $$T$$:

$$H_\beta-H_\alpha=\int_{0}^{T}\Delta C_P\,dT=\Delta C_P\;T=1\times T=T \quad\text{J mol}^{-1}$$

At the transition temperature $$T_t=600\ \text{K}$$, the two phases are in equilibrium, so $$\Delta G=0$$ and

$$\Delta H_{t}=T_t\,\Delta S_{t}\;.$$

Hence the entropy difference at 600 K is

$$\Delta S_{600}=S_\beta-S_\alpha =\frac{\Delta H_{t}}{T_t} =\frac{600\ \text{J mol}^{-1}}{600\ \text{K}} =1\ \text{J mol}^{-1}\,\text{K}^{-1}$$

To obtain the entropy difference at 300 K, integrate from 300 K to 600 K (using the given constant $$\Delta C_P$$):

$$\begin{aligned} \Delta S_{300} &=\Delta S_{600}-\int_{300}^{600}\frac{\Delta C_P}{T}\,dT\\[4pt] &=1-\int_{300}^{600}\frac{1}{T}\,dT\\[4pt] &=1-\left[\ln T\right]_{300}^{600}\\[4pt] &=1-\ln\!\left(\frac{600}{300}\right)\\[4pt] &=1-\ln 2\\[4pt] &=1-0.69\\[4pt] &=0.31\ \text{J mol}^{-1}\,\text{K}^{-1} \end{aligned}$$

Therefore, the entropy change at 300 K is

0.31 J mol-1 K-1.