For the general case, you should consider the base 10 logarithm of the number. It should be easy to simplify things since $\log_{10}{ab} = \log_{10}{a} + \log_{10}{b}$.
But since calculators are not allowed for your specific example, we'll have to do it a bit differently. The product in your specific example is roughly equals to $4 \times 50 \times 6000 \times 70000 \times 800000 \times 9000000$. I guess its easier to calculate the number of digits from here on, and probably the only way (AFAIK) to do it without a calculator for your specific example.
Write $a,b,c$ for the sums of each of the three rows. We have $a+b+c=1+\cdots+9=45$. The fact that each row is a multiple of 9 means that $a,b,c\in\{9,18,27,36\}$. So $a,b,c$ are either $27,9,9$ or $18,18,9$, in some order. As $1+2+3+4=10$, the 4-digit term cannot be $9$, so it is $18$ or $27$. But $27$ is not the sum of four consecutive integers, so the third row is $3,4,5,6$ in some order.
We are left with $1,2,7,8,9$ for the other two rows, one adding to $18$ and the other $9$. We cannot reach $18$ with two digits at all or with three digits not using the $9$, so the first row is necessarily one of the following two possibilities: $9,8,1$ in some order, and the middle one $2,7$; or $2,7,9$ in some order and $1,8$. This last case is easily discarded since the only 4-digit product is $279\times18=5022$, which produces the wrong last row.
Looking at the first row, the last digit cannot be $1$ because that would imply a repeated digit. $2$ and $9$ do not fit, because they would put an $8$ in the last row. Actually, if the first row starts with $8$ or $9$, as $819\times27$ already has five digits, we learn that the first number starts with $1$ and that the second one is $27$.
We have $189\times27=5103$, not good, so it has to be $198\times 27=5346$. So the answer to the question is $8+7+6=21$.
Best Answer
You are absolutely correct, and what you did is perhaps a more intuitive way of the logarithm-based method.
When you say $a$ has 19 digits, you are inherently saying its logarithm is $18.xxx$, and similarly for $b=8.yyy$. Thus $ab$ has either $27$ or $28$ digits ($\log_{10}ab=26+z$ where $z<2$), but when you upper-bound both numbers and show that the product of the first digits is less than $10$, you are inherently saying that $.xxx+.yyy<1$, i.e. the numbers are not big enough to introduce an additional digit. Hence there are $27$ digits.