Answer
Centred at $2$.
Work Step by Step
Re-write the given series in the expanded form as follows:
$\Sigma_{n=1}^\infty \dfrac{(x-2)^n}{n^3}=\dfrac{(x-2)}{1^3}+\dfrac{(x-2)^2}{2^3}+\dfrac{(x-2)^3}{3^3}+.....$
We can see that the above power series is centred at $2$.
