The elements ${{ }^{\text{'}}}^{\mathrm{\prime }}{\mathrm{A}}^{\text{'}}$ and ${ }^{\text{'}}{\mathrm{B}}^{\text{'}}$ combine together to give two  compounds ${\mathrm{A}}_2{\mathrm{B}}_3$ and ${\mathrm{AB}}_2$. The weight of 0.2 mole of ${\mathrm{A}}_2{\mathrm{B}}_3$ is$26\mathrm{g}\mathrm{. }$ . The weight of 0.3 mole of ${\mathrm{AB}}_2$ is $24\mathrm{g}.$  Then the  atomic weights of   $\mathrm{A}$and  $\mathrm{B}$ respectively

Given, The Weight of  of
${\mathrm{A}}_2{\mathrm{B}}_3$ is $26\mathrm{g}$ The weight of  of ${\mathrm{AB}}_2$  is  $24\mathrm{g}$
Atomic weight of  A and  B = ? Weight of 0.2 mole of  $A_2{B}_3\to 26\mathrm{g}$
Weight of $\to$? $\frac{1\times 26}{0.2}=130\mathrm{g}$one mole  ?  Weight of 0.3 mole of $1B_2\to 24\mathrm{g}$
Weight of 1 mole of $A\to ?\frac{1\times 24}{0.3}=80\mathrm{g}$ 
$$\begin{gathered} 2\mathrm{A}+3\mathrm{B}\to 130\dots \dots .\left(1\right) \\ \mathrm{A}+2\mathrm{B}\to 80 \end{gathered}$$ 
On Solving : $2\mathrm{A}+4\mathrm{B}\to 160$
$160-4B+3B = 130$

B = 30

2A = 130 - 90 = 40

A = 20