jmc

Since each coin has 2 possible outcomes, there are $2^n$ possible outcomes for the $n$ coins. The number of outcomes in which the number of tails is 0 or 1 is $\left(\genfrac{}{}{0ex}{}{n}{0}\right)+\left(\genfrac{}{}{0ex}{}{n}{1}\right)=1+n$. So the probability of having at most one tail is $\frac{1+n}{2^n}$. Therefore, we must solve the equation $$\frac{1+n}{2^n}=\frac{3}{16}.$$ We can check (simply by plugging in values of $n$) that if $1\le n\le 5$, then $n=5$ is the only solution. Now we show that $n\ge 6$ cannot be a solution to the equation. Observe that $n\ge 6$ implies $n<2^{n-3}$, thus $$\frac{1+n}{2^n}<\frac{1+2^{n-3}}{2^n}=\frac{1}{2^n}+\frac{1}{8}<\frac{1}{16}+\frac{1}{8}=\frac{3}{16}.$$ So there are $\boxed{5}$ coins.