$0.01$ mole of iodoform(${\mathrm{CHI}}_3$) reacts with Ag powder to produce a gas Whose volume at NTP is        

Balanced equation is as follows

           $2{\mathrm{CHI}}_3+6\mathrm{Ag}\to {\mathrm{C}}_2{\mathrm{H}}_2+6\mathrm{AgI}$ 

$2$ moles of Iodoform makes $1$ mole of acetylene, so $0.01$ mole will form $0.005$ moles of acetylene.

Volume occupied by one mole gas at NTP is $22.4$ Liters.

Hence volume occupied by $0.005$ mole will be $22.4 \times 0.005=0.112$ Liters.$0.112\mathrm{Litres}=112\mathrm{ml}$.