Answer
33 g
Work Step by Step
Heat gained by the ice=$mL_{F}+mc\Delta T=m(L_{F}+c\Delta T)$
$=.150\,kg(333\,kJ/kg+4.1868\,kJ/kg\cdot ^{\circ}C\times(50-0)^{\circ}C$)
$=81.351\,kJ$
Heat lost by the steam=$mL_{V}+mc\Delta T=m(L_{V}+c\Delta T)$
$=m_{steam}(2256\,kJ/kg+4.1868\,kJ/kg\cdot ^{\circ}C\times50^{\circ}C)$
($L_{V}$ and $\Delta T$ will be negative, but we neglect the negative sign as it only indicates that the heat is lost)
Now, Heat lost by the steam= Heat gained by the ice.
Therefore,
$m_{steam}(2256\,kJ/kg+4.1868\,kJ/kg\cdot ^{\circ}C\times50^{\circ}C)=81.351\,kJ$
$\implies m_{steam}=\frac{81.351\,kJ}{2256\,kJ/kg+4.1868\,kJ/kg\cdot ^{\circ}C\times50^{\circ}C}=0.033\,kg=33\,g$
