Answer: B

Solution:

The sum of the logarithms is an integer $k$ exactly when the product of the six numbers is $10^k.$ Since $10 = 2 \cdot 5,$ each chosen number must be of the form $2^a 5^b,$ so it comes from $$1, 2, 4, 5, 8, 10, 16, 20, 25, 32, 40.$$

For each, record the excess of factors of $2$ over factors of $5$: $0, 1, 2, -1, 3, 0, 4, 1, -2, 5, 2.$ The product is a power of $10$ only if the six chosen values have equal totals of $2$s and $5$s, i.e. their excesses sum to $0.$

The only negative excesses are $-1$ for $5$ and $-2$ for $25.$ A six-number ticket with total excess $0$ must contain both: omitting either leaves too few zero and small positive excesses to reach six numbers. The other four numbers must therefore have total excess $3.$ There are two numbers of excess $0$ ($1,10$), two of excess $1$ ($2,20$), and two of excess $2$ ($4,40$); any number of excess at least $3$ would leave too few zeros to complete a four-number selection. Thus we must take both excess-$0$ numbers, one excess-$1$ number, and one excess-$2$ number. This gives exactly four valid tickets: $\{1, 5, 10, 20, 25, 40\},$ $\{1, 2, 5, 10, 25, 40\},$ $\{1, 2, 4, 5, 10, 25\},$ and $\{1, 4, 5, 10, 20, 25\}.$

Professor Gamble holds one of these four, and only one matches the winning ticket, so the probability is $\dfrac14.$

Thus, the correct answer is B.