Let distances from the lighthouse be $x$ (for $30^\circ$) and $y$ (for $45^\circ$). With height $h=100$ m:

$\tan 30^\circ=\dfrac{100}{x}\ \Rightarrow\ x=\dfrac{100}{\tan30^\circ}=100\sqrt{3}$,   $\tan 45^\circ=\dfrac{100}{y}\ \Rightarrow\ y=\dfrac{100}{\tan45^\circ}=100$.

Ships are on opposite sides ⇒ distance $=x+y=100\sqrt{3}+100=100(\sqrt{3}+1)\approx \boxed{273.2\ \text{m}}$.