Answer
The mass of the steam required is 0.180 kg
Work Step by Step
We can find the heat energy required to melt the ice and raise the temperature up to $30^{\circ}C$.
$Q = m~L + m~c_w~\Delta T$
$Q = (1.00~kg)(3.33\times 10^5~J/kg) + (1.00~kg)(4186~J/kg~C^{\circ})(30~C^{\circ})$
$Q = 458,580~J$
The heat energy required by the ice will be equal to the heat energy lost by the steam.
$Q = m~L+m~c_w~\Delta T$
$m = \frac{Q}{L+c_w~\Delta T}$
$m = \frac{458,580~J}{(2.26\times 10^6~J/kg) + (4186~J/kg~C^{\circ})(100~C^{\circ}-30~C^{\circ})}$
$m = 0.180~kg$
The mass of the steam required is 0.180 kg.