jmc

We claim that $a$ is a perfect $n$th power if and only if $n$ divides both $306$ and $340$. To see this, suppose that $n\mid 306$ and $n\mid 340$. Then $2^{\frac{306}{n}}{3}^{\frac{340}{n}}$ is an integer whose $n$th power is $a$. Conversely, suppose $b^n=a$. Then the only primes which divide $b$ are $2$ and $3$. Choose $c$ and $d$ so that $b=2^c{3}^d$. Then $b^n=2^{cn}{3}^{dn}=2^{306}{3}^{340}$, which implies $n\mid 306$ and $n\mid 340$. This concludes our proof of the claim that $a$ is an $n$th power if and only if $n$ divides both $306$ and $340$.

The largest number which simultaneously divides two numbers is their GCD. Using the Euclidean algorithm, the GCD of $306$ and $340$ is the same as the GCD of $340$ and $340-306=34$. Since $34$ divides $340$, the GCD of these two is $34$, so the largest possible $n$ is $\boxed{34}$.