Correct Answer: $1000\sqrt{3}$

Solution :
Let D be the aeroplane.
A and B are two ships.
Given:
$CD = 1500 \text{ m,} \angle EDA = 30^{\circ}, \angle EDB = 60^{\circ}$
In $\triangle DBC,$
$\tan 60^{\circ} = \frac{DC}{BC}$
$\Rightarrow \sqrt{3} = \frac{1500}{BC}$
$\Rightarrow BC = \frac{1500}{\sqrt{3}} \text{ m}$
In $\triangle ADC, $
$\tan 30^{\circ} = \frac{DC}{AC}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{1500}{AB + BC}$
$\Rightarrow AB + BC = 1500\sqrt{3}$
$\Rightarrow AB = 1500\sqrt{3} - \frac{1500}{\sqrt{3}}$
$\Rightarrow AB = 1500(\sqrt{3} - \frac{1}{\sqrt{3}})$
$\Rightarrow AB = 1500( \frac{3-1}{\sqrt{3}})$
$\Rightarrow AB = \frac{3000}{\sqrt{3}}$
$\Rightarrow AB = 1000\sqrt{3}$ metres.
Hence, the correct answer is $1000\sqrt{3}$ metres.