b
$f = 6$
$\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{6} = \frac{4}{3}$
$\frac{{\Delta W}}{{\Delta Q}} = \left( {1 - \frac{1}{\gamma }} \right)$
==> $\frac{{25}}{{\Delta Q}} = \left( {1 - \frac{1}{{4/3}}} \right)$
$ = 1 - \frac{3}{4} = \frac{1}{4}$
==> $\Delta Q = 25 \times 4 = 100\,Joule$