Explanation

(d) Suppose $m\ gm$ ice melted, then heat required for its melting $= mL = m \times 80\ cal$
Heat available with steam for being condensed and then brought to $0℃$
$= 1 \times 540 + 1 \times 1 \times (100 - 0) = 640\ cal$
$\Rightarrow$ Heat lost = Heat taken
$\Rightarrow 640 = m \times 80 \Rightarrow m = 8\ gm$
Short trick : You can remember that amount of steam $(m')$ at $100℃$ required to melt $m\ gm$ ice at $0℃$ is
$m^{'} = \frac{m}{8}$
Here, $m = 8 \times m' = 8 \times 1 = 8\ gm$