Average atomic weight of  Boron is 10.10 and boron exists in two isotopic forms $B^{10}and{B}^{11}$. The percentage abundance of $B^{10}$ is

Assume percentage abundance of  $B^{10}=x$
So percentage abundance of  $B^{11}=100-x$
Apply formula,  $\frac{x\times 10+(100-x)11}{100}=10.10$
 $$\begin{gathered} \Rightarrow 10-1100-11x=1010 \\ \Rightarrow x=90\% \end{gathered}$$