jmc

Recall the identity $\mathrm{lcm}[a,b]\cdot \gcd (a,b)=ab$, which holds for all positive integers $a$ and $b$. Applying this identity to $12$ and $t$, we obtain $$\mathrm{lcm}[12,t]\cdot \gcd (12,t)=12t,$$and so (cubing both sides) $$\mathrm{lcm}[12,t{]}^3\cdot \gcd (12,t{)}^3=(12t{)}^3.$$Substituting $(12t{)}^2$ for $\mathrm{lcm}[12,t{]}^3$ and dividing both sides by $(12t{)}^2$, we have $$\gcd (12,t{)}^3=12t,$$so in particular, $12t$ is the cube of an integer. Since $12=2^2\cdot 3^1$, the smallest cube of the form $12t$ is $2^3\cdot 3^3$, which is obtained when $t=2^1\cdot 3^2=18$. This tells us that $t\ge 18$.

We must check whether $t$ can be $18$. That is, we must check whether $\mathrm{lcm}[12,18{]}^3=(12\cdot 18{)}^2$. In fact, this equality does hold (both sides are equal to $6^6$), so the smallest possible value of $t$ is confirmed to be $\boxed{18}$.