The total heat change, $ \Delta H = 90 \times 540 = 48600 \: cal$
Now, $ \Delta H = \Delta E + P\Delta V$
$\Delta V = (V_{vapor} – V_{liquid}) = V_{vapor} (V_{liquid} is negligible)$
$\Delta H = \Delta E + PV_{vapor} = \Delta E + nRT$
$\Rightarrow \Delta E = \Delta H – nRT$
$ = 48600 – 90/18 \times 2 \times 373 = 48600 – 3730 = 44870 \: cal$
Ans : (A)