Answer
$3.4\times 10^{14}Hz$
Work Step by Step
We can find the required frequency as
$f_{peak}=(5.88\times 10^{10}s^{-1}K^{-1})T$
We plug in the known values to obtain:
$f_{peak}=(5.88\times 10^{10}s^{-1}K^{-1})(5800K)$
$f_{peak}=3.4\times 10^{14}Hz$
