Does base matter?

$30$ in base $n$ is equal to $3n$ .

To get an integer $n$ , $3n$ has to be equal to a number divisible by $3$ (as well as by $5$ as stated in the problem).

$3n$ needs to be larger than $30$ , since all potential bases are larger than $10$ .

The first number divisible by both $5$ and $3$ and larger than $30$ is $45$ .

$3n=45$ from which $n=\boxed{15}$

5 is divisible by numbers that ends with 0 or 5. Clearly from the options, 5 < 11, 12, 13, 14, 15. Therefore 5 remains 5 when converted to base ten from any of the options given. However, 30 > 11, 12, 13, 14, 15. Therefore converting to base ten would involve expansion by the powers of base. Hence base15 as it is the only option that gives a number that ends with 5, which is therefore divisible by 5.

We note that $\dfrac {30_b}{5_b} = \dfrac {3_{10}b_{10}}{5_{10}}$ , where $b$ denotes the number base. Since $b >5$ , $5_b = 5_{10}$ . $30_b$ is divisible by $5_b$ only if $b$ is divisible by 5, therefore, it is divisible when $b=5, 10, \boxed{15},…$