The percentage of two stable boron isotopes ${}_5{B}^{11}$ and ${}_5{B}^{10}$ is, given that average atomic mass of boron is 10.81amu.

The correct option is A 81.0 % and 19.0 %
Average atomic mass of Boron $=10.81$ amu
Let $x$ be the % of $B^{11}$.

Hence,
$10.81=\frac{x\times 11+(100-x)10}{100}\Rightarrow x=81$