Solution:

The number of pennies must have the same remainder as $102$ modulo $5$, so there are either $2$ or $7$ pennies. Seven pennies would leave only two coins for nickels, dimes, and quarters, impossible because at least one of each type is needed.

So there are $2$ pennies. The remaining $7$ coins are worth $100$ cents. If $n,d,q$ are the numbers of nickels, dimes, and quarters, then $n+d+q=7$ and $5n+10d+25q=100$.

Dividing the value equation by $5$ and subtracting the coin-count equation gives $d+4q=13$. The only positive solution is $q=3$, $d=1$, and $n=3$.

Thus there must be $1$ dime.

Thus, A is the correct answer.