At 0°C ice and water are in equilibrium and $∆\mathrm{G} \mathrm{and} ∆\mathrm{S}$  for the conversion of ice to liquid water is:

Since the given process is in equilibrium i.e., 

$\begin{array}{l}\mathrm{\Delta G}=0\\ \mathrm{\Delta G}=\mathrm{\Delta H}-\mathrm{T\Delta S}\end{array}$

or

$$\begin{gathered} \mathrm{O}=\mathrm{\Delta H}-\mathrm{T\Delta S} \\ \mathrm{\Delta H}=\mathrm{T\Delta S} \\ \mathrm{\Delta S}=\frac{\mathrm{\Delta H}}{\mathrm{T}} \end{gathered}$$

Substituting the values:

$$\begin{gathered} ∆\mathrm{S}=\frac{6}{273}\times {10}^3 \mathrm{J} {\mathrm{K}}^{-1}{ \mathrm{mol}}^{-1} \\ =21.98 \mathrm{J} {\mathrm{K}}^{-1}{ \mathrm{mol}}^{-1} \end{gathered}$$