jmc

Let $abc$ represent the three-digit number in base 3, where $a$, $b$ and $c$ each represent a digit 0, 1 or 2. The place values in base 3 are 9, 3 and 1, so the base-ten value of $abc$ is $a\times 9+b\times 3+c\times 1$, which can be written as $9a+3b+c$. This same value is $cba$ in base 4, which we can write as $16c+4b+a$. Equating these two expressions, we get $9a+3b+c=16c+4b+a$. We can simplify this to $8a=15c+b$. Now, there are only three digits to try for each letter. It turns out that $8\times 2=15\times 1+1$, so the base-three number is $211_3$ and the base-four number is $112_4$. The base-ten value is $(2\times 9)+(1\times 3)+1=18+3+1=22$. To confirm this answer, we check the base-four value: $1\times 16+1\times 4+2\times 1=16+4+2=\boxed{22}.$