smc

Looking at the prime numbers under $12$, we see that there are $\lfloor \frac{12}{2}\rfloor +\lfloor \frac{12}{2^2}\rfloor +\lfloor \frac{12}{2^3}\rfloor =6+3+1=10$ factors of $2$, $\lfloor \frac{12}{3}\rfloor +\lfloor \frac{12}{3^2}\rfloor =4+1=5$ factors of $3$, and $\lfloor \frac{12}{5}\rfloor =2$ factors of $5$. All greater primes are represented once or none in $12!$, so they cannot be part of the square. Since we are looking for a perfect square, the exponents on its prime factors must be even, so we can only use $4$ of the $5$ factors of $3$. Therefore, the prime factorization of the square is $2^{10}\cdot 3^4\cdot 5^2$. To find the square root of this, we halve the exponents, leaving $2^5\cdot 3^2\cdot 5$. The sum of the exponents is $\boxed{8}$