Solution

Given,
$$S_{\infty} = \dfrac{a}{1-r} = 162\quad...(i)\\S_n = \dfrac{a(1-r^n)}{1-r} =160\quad...(ii)$$

Dividing equation $$(ii)$$ by $$(i)$$,
$$\ \ 1 - r^n = \dfrac{160}{162} = \dfrac{80}{81}\\ \Rightarrow 1 - \dfrac{80}{81} = r^n \\ \Rightarrow r^n = \dfrac{1}{81} \\ \Rightarrow {\left( \dfrac{1}{r} \right)}^n = 81$$

Now, it is given that $$\dfrac 1r$$ is an integer and $$n$$ is also an integer.
Hence, $$ \dfrac 1r = 3, 9 , 81$$ so that $$n = 4, 2 , 1$$.
$$\therefore$$ $$a = 162 \left( 1 - \dfrac{1}{3} \right)$$ or $$162 \left( 1 - \dfrac{1}{9} \right)$$ or $$162 \left( 1 - \dfrac{1}{81} \right)$$
$$\qquad = 108 $$ or $$144 $$ or $$160$$

Hence, the option (D) is correct.