The series of positive multiples of 3 is divided into sets : 3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 ,…

Given, $\left\{3\right\},\left\{6,9,12\right\},\left\{15,18,21,24,27\right\},\cdots$

Now, the number of elements in ${11}^{\text{th }}$ set will be $=1+\left(10\right) 2=21$ 

The total number of elements up to ${10}^{\text{th}}$ set will be $1+3+...+19$$={10}^2=100$

$\therefore$ elements in ${11}^{\text{th}}$ set $=\left\{3·101,3·102, ....3·121\right\}$

Sum of these elements $=3\left(101+102+\cdots +121\right)$

$=3\times \frac{21}{2}\times \left(101+121\right)=6993$