Number of moles of alcoholic compound $=\frac{\mathrm{Given} \mathrm{mass}}{\mathrm{Molar} \mathrm{mass}} = \frac{1.84\times {10}^{-3}}{92.0} = 2\times {10}^{-5} \mathrm{moles}$

Number of moles of hydrogen gas produced $= \frac{\mathrm{Volume} \mathrm{of} \mathrm{gas} \mathrm{at} \mathrm{STP}}{22400 \mathrm{ml}} = \frac{1.344}{22400} = 6\times {10}^{-5} \mathrm{moles}$

Monohydric alcohol will react with active metal to give $\frac{1}{2} \mathrm{mole}$ of hydrogen gas.

Since per mole of given alcohol is giving 3 moles of hydrogen gas.

So number of alcoholic hydrogen present in compound should be =6.