At $pH =1$
Concentration of $\left[ H ^{+}\right]=10^{-1} M =0.1\, M$
At $pH =2$
Concentration of $\left[ H ^{+}\right]=10^{-2} M =0.1\, M$
Now
$M _{1} V _{1}= M _{2} V _{2}$
$0.1\, M \times 11=0.01\, M \times V _{2}$
$V _{2}=10\, L$
Thus the final volume of solution is $10\, L$. Hence $9\, L$ should be added to $1\, L$ of solution to
make $pH =2$ from $pH =1$