Incomparable numbers!

$n^{20} > 5^{30}$

$(n^2)^{10} > (5^3)^{10}$

$n^2 > 5^3$ $\Rightarrow$ $n^2 > 125$ So, $n = 12$ .

$n^{20}=(n^2)^{10}>5^{30}=(5^3)^{10}$

From that $n^2>5^3$ has to be true, so the answer is $\boxed{12}$ , because $11^2=121<5^3=125<12^2=144$

$n^{20}>5^{30}$

$n^{20}>(5^{20})^{1.5}$

$n>5^{1.5}$

$5^{1.5}\approx 11$

For $n$ to be as small as possible (and remain an integer) it must be 1 bigger than 11, which is $\boxed{12}$ .