That is an interesting question. I'm not yet sure of a good way to solve the problem. Here are my thoughts.
Off the bat, since 7<10, the number of digits in 7^54 must be somewhat less than 54. Also since 7 is greater than the square root of 10 (which is somewhere between 3 and 4) then the number of digits in 7^54 is greater than 27. Along the same vein, 7 is greater than 10^(2/3) (which is between 4 and 5) so 7^54 must have more than 36 digits. For a closer estimate, I'd do the following calculation:
The prime factorization of 54 is
, so we can write
Here we say that 49 is pretty close to 50 and compute a bit. The cube of 50 is easy, and we estimate that
We need to cube this value twice more. The 10s are easy to cube, and we have
You can use the binomial approximation to estimate that 1.25^3 is about 1.75, which is less than 2. The cube of 2 is less than ten, so 1.25^9 is less than 10 and does not increase the number of digits. So 7^54 has about 45 digits. Since we used 50 to approximate 49, this is an upper bound.
It turns out that
so this upper bound is very sharp.
There will be some problems on the test which can actually be solved by clever estimation. It is a multiple choice test, and you can use this to your advantage.
[EDIT]: In retrospect, this method might not be so bad. The overall plan is to multiply everything out, and when things start to get complicated, make estimates to make your computation easier. By using the binomial approximation
you can get a good first order approximation of your error terms. Since you are going for order of magnitude, this should be good enough.