For the following reaction scheme, percentage yields are given along the arrow: $\mathbf{x} g$ and…

Now, one mole of propyne give, one mole of $2$-butyne. $3$ moles of $2$-butyne give one mole of hexamethyl benzene.

Hence, $40 \mathrm{g}$ of propyne give $54 \mathrm{g}$ to $2$-butyne $\left(100\%\right)$ but given that $75\%$ yeild and $4 \mathrm{g}$ to propyne.

The mass of $2$-butyne formed $=5.4\times \frac{75}{100}=4.05 \mathrm{g}$ $162 \mathrm{g}$ to $2$-butyne give $162 \mathrm{g}$ of hexamethyl benzene $\left(100\% \mathrm{yield}\right)$

But given that $40\%$ yield, and $4.05 \mathrm{g}$ of $2$-butyne

The mass of hexamethyl benzene formed $=4.05\times \frac{40}{100}=1.62 \mathrm{g} \left(\mathrm{x}\right)$