Answer
The speed of the train is 25.0 m/s.
Work Step by Step
Let $F_T$ be the tension in the cord holding the lamp.
$m\frac{v^2}{r} = F_T ~sin(\theta)$
$mg = F_T ~cos(\theta)$
We can divide the first equation by the second equation.
$\frac{v^2}{gr} = tan(\theta)$
$v = \sqrt{gr ~tan(\theta)} = \sqrt{(9.80 ~m/s^2)(215 ~m)(tan(16.5^{\circ}))}$
$v = 25.0 ~m/s$
The speed of the train is 25.0 m/s.
