This is how an full adder is implemented using XOR gate, AND and OR gates.
$C_{out} = (A\oplus B)C + AB$
We compute $A \oplus B$ and $AB$ parallely here, So time taken to compute $A \oplus B$ and $AB$ is equal to the maximum of time taken to compute these two individually = $max(A \oplus B,AB) = 30ns$
Now, We find $ (A\oplus B)C $ next which takes additional $10ns$ due to AND gate and next we OR $(A\oplus B)C$ and $AB$, which takes additional $20ns$ time.
So, total time taken to find out Carry $= 30 + 10 +20 = 60ns$