A and B are two elements which form $A{B}_{2}and{A}_2{B}_3$ weights 17.8g. Then 

0.18 mole of  $A{B}_2$ weighs 10.6 g

1 mole of $A{B}_2$ weights $\frac{1\times 0.6}{0.18}=58.88g$

0.18 mole of $A_2{B}_3$ weighs 17.8 g

1 mole  $A_2{B}_3$ weighs $\frac{1\times 17.8}{0.18}=98.88g$ 

$A{B}_2:x+2y=58.88g\mathrm{....}\left(1\right)$

$A_2{B}_3:2x+3y=98.88g\mathrm{.....}\left(2\right)$

$1\times 22x+4y=117.76$

$-23x+3y=98.88$            

   $y=18.88g$         

   $\therefore$ Atomic mass of B = 18.88 g/ mol

Atomic mass of A = x + 2 (18.88) = 58.88

                                       x = 58.88 – 37.76 = 21.12 g/mol