Answer
$\Delta S=+943\frac{J}{K}$
Work Step by Step
$m_{i1}+m_{i2}=m_{f1}+m_{f_2}$
Since $m_{f1}=m_{f2}$
$\implies m_{i1}+m_{i2}=2m_{f1}$
This can be rearranged as:
$m_{f1}=\frac{m_{i1}+m_{i2}}{2}$
$m_{f1}=\frac{1.773+0.227}{2}=1Kg$
Now, $m=m_{i1}-m_{f1}=1.773-1=0.773Kg$
$0.773Kg$ of ice must be melted that is why heat must be added so heat is positive.
We also know that:
$Q=mL_F$
We plug in the known values to obtain:
$Q=(0.773)(333000)=257409J$
Now,
$\Delta S=\int _{Q_i} ^{Q_f}\frac{dQ}{T}=\frac{Q}{T}$
We plug in the known values to obtain:
$\Delta S=\frac{257409}{273}=+943\frac{J}{K}$