If a

LCM $(a, b, c)=2^{3} \times 3^{2} \times 5 \ldots \ldots$ (I)

We have to find the value for n

Also

$a=2^{3} \times 3$

$b=2 \times 3 \times 5$

$c=3^{n} \times 5$

We know that the while evaluating LCM, we take greater exponent of the prime numbers in the factorization of the number.

Therefore, by applying this rule and taking $n \geq 1$ we get the LCM as

$\operatorname{LCM}(a, b, c)=2^{3} \times 3^{n} \times 5 \ldots \ldots$ (II)

On comparing (I) and (II) sides, we get:

$2^{3} \times 3^{2} \times 5=2^{3} \times 3^{n} \times 5$

$n=2$

Hence the correct choice is (b).