If $n$ is a positive integer such that $n^2$ is divisible by $72$, then the smallest possible value of $n$ is:

Step 1 (Factorize 72): We find the prime factorization of 72:

Step 2 (Condition for smallest n): For $n^2$ to be divisible by $72$, the exponents of the primes in $n^2$ must be at least as large as those in 72:

• Exponent of 2 in $n^2$ must be $\ge 3$. Since the exponent in a perfect square is always even, the exponent of 2 in $n^2$ must be at least $4$. This implies $n$ must be divisible by $2^2=4$.

• Exponent of 3 in $n^2$ must be $\ge 2$. This implies $n$ must be divisible by $3^1=3$.

Step 3 (Compute smallest n): The smallest positive integer $n$ is:

Step 4 (Conclusion): The smallest value of $n$ is 12, which is at option index 1 in the list ['6', '12', '18', '24'].