(c)
$\because Mass\ of\ 22400\ {cm}^{3}\ of\ {CH}_{4}\ at\ STP = 16\ g\$
$\therefore\ \ \ \ \ \ \ Mass\ of\ 1\ {cm}^{3}of\ {CH}_{4}\ at\ STP = \ \frac{16}{22400}g$
$\therefore Mass\ of\ 112\ {cm}^{3}\ of\ {CH}_{4}\ at\ STP = \frac{16}{22400}\ \times 112$
= 0.08 g