$1\, g$ of a compoun react with excess of $CH_3MgI$  to give $\frac{x \times 22400}{92}$

$0.092$ of a compoun react with excess of $CH_3MgI$  to give $\frac{x \times 22400}{92} \times 0.092$

$\frac{x \times 22400 \times 0.092}{92} = 67\, mL$ of $CH_4$ at $STP$

$x = \frac{67 \times 92}{22400 \times 0.092} = \frac{67 \times 92 \times 1000}{22400 \times 92} = \frac{670}{224} = 2.99 \approx 3$

$x \Rightarrow$ number of active hydrogen in compound. active $H  \Rightarrow 3$