Given, mass of steam, $m_\lambda=5 \mathrm{~g}$
Temperature of steam, $T_s=100^{\circ} \mathrm{C}$
Mass of ice, $m_i=5 \mathrm{~g}$
Temperature of ice, $T_i=0^{\circ} \mathrm{C}$
Since, latent heat of vaporisation, $l_v=540 \mathrm{cal} / \mathrm{g}$
Latent heat of fusion, $l_f=80 \mathrm{cal} / \mathrm{g}$
Now, heat given by steam during fully conversion into water, at $100^{\circ} \mathrm{C}$
and
$$
Q_1=m_\lambda \times 540=2700 \mathrm{cal}
$$

Heat taken by ice to get fully converted into water at $100^{\circ} \mathrm{C}$,
$$
\begin{aligned}
Q_2 & =m_i l_f+m_i \times s \times \Delta T \\
& =5 \times 80+5 \times 1 \times 100 \quad(\because s=1, \text { for water }) \\
& =400+500=900
\end{aligned}
$$

As, $Q_2 < Q_1$
$\therefore$ Temperature of mixture will be $100^{\circ} \mathrm{C}$.