Answer
$\Delta S=14.64\frac{J}{K}$
Work Step by Step
We can find the entropy change as
$\Delta S=\frac{Q}{T}$......eq(1)
Also,
$Q=mL_f$ where $L_f$ is water heat of fusion and its value is $333000\frac{J}{kg}$.
Thus, $Q=(0.012)(333000)=3996J$
We plug in the known values in eq(1) to obtain:
$\Delta S=\frac{3996}{273}=14.64\frac{J}{K}$