Given that, speed of the train $S = 60\;km/hr = 60 \times \dfrac{1000}{60 \times 60}\;m/sec = \dfrac{50}{3}\;m/sec$
and, time to cross the pole $T =9\;sec $
Let’s lenth of the train $ = D\;m$
We know that $S = \dfrac{D}{T}$
$\implies D = S \times T$
$\implies D = \dfrac{50}{3}\;m/sec \times 9\;sec$
$\implies D = 150\;m.$
So, the correct answer is $(D).$