jmc

Let $p$ and $q$ be the prime divisors of $n$, so we can write $n=p^a\cdot q^b$ for positive integers $a$ and $b$. This means $n^2=p^{2a}\cdot q^{2b}$, so $t(n^2)=(2a+1)(2b+1)=27$. Since $2a+1$ and $2b+1$ are both greater than 1 and are divisors of 27, we know they are 3 and 9 (in no particular order). This means that $a$ and $b$ are 1 and 4 (in no particular order), so $$t(n)=(a+1)(b+1)=(1+1)(4+1)=\boxed{10}.$$