(c) $Surface\ energy = surface\ tension \times surface\ area\ $
$E = T \times 2A$
New surface energy,
$E_{1} = T \times 2\left( \frac{A}{2} \right) = T \times A$
$\%\ decrease\ in\ surface\ energy = \frac{E - E_{1}}{E} \times 100$
$\ = \frac{2TA - TA}{2TA} \times 100 = 50\%$