Molar mass of benzene =$78 \mathrm{g}/\mathrm{mol}$

Mole of benzene = $\frac{\mathrm{Given} \mathrm{mass}}{\mathrm{Molar} \mathrm{mass}}=\frac{7.8}{78}=0.1 \mathrm{mol}$

${\mathrm{C}}_6{\mathrm{H}}_6 + {\mathrm{CH}}_3\mathrm{COCl}/{\mathrm{AlCl}}_3\to {\mathrm{C}}_8{\mathrm{H}}_8\mathrm{O}$

By applying the mole-mole analysis, one mole of benzene gives the one mole of ${\mathrm{C}}_8{\mathrm{H}}_8\mathrm{O}$. So, $0.1 \mathrm{mol}$ of benzene gives the $0.1 \mathrm{mol}$ of ${\mathrm{C}}_8{\mathrm{H}}_8\mathrm{O}$. 

Molar mass of ${\mathrm{C}}_8{\mathrm{H}}_8\mathrm{O}$= $120 \mathrm{g}/\mathrm{mol}$

Mass of ${\mathrm{C}}_8{\mathrm{H}}_8\mathrm{O}$ formed = $\mathrm{mole}\times \mathrm{molar} \mathrm{mass}=0.1\times 120=12\mathrm{g}$

Percentage yield =$\frac{\mathrm{Experimental} \mathrm{yield}}{\mathrm{Theoritical} \mathrm{yield}}\times 100=\frac{8.4}{12}\times 100=70$

So, the percentage yield of the product ${\mathrm{C}}_8{\mathrm{H}}_8\mathrm{O}$ is $70\%$.