You're either asking to evaluate
1101² + 1101²
or, seeing as 1101 consists only of 0s and 1s, you're asking to compute a sum in base 2,
1101₂ + 1101₂
If you really mean 1101² + 1101², we have
1101² = (1100 + 1)²
… = 1100² + 2•1100•1 + 1²
… = 1,210,000 + 2200 + 1
… = 1,212,201
so that
1101² + 1101² = 2 • 1,212,201 = 2,424,402
I think the latter is the more likely scenario. In base 2, we have
0₂ + 0₂ = 0₂
1₂ + 0₂ = 0₂ + 1₂ = 1₂
1₂ + 1₂ = (2)₂ = (2 + 0)₂ = 10₂
So, when we add the digits of 1101₂ and 1101₂ in the same place value, we have
• in the "ones" place, 1₂ + 1₂ = 10₂
• in the "twos" place, 0₂ + 0₂ + 1₂ = 1₂, where we carry the leading 1 from the previous place sum
• in the "fours" place, 1₂ + 1₂ = 10₂
• in the "eights" place, 1₂ + 1₂ + 1₂ = 11₂, where we again carry the leading 1 from before
Then
1101₂ + 1101₂ = 11010₂