For a general $n$, the constant term in $$f(x) = (1 + x/5)^n (2-3/x)^2$$ can be found by observing first that $$(2-3/x)^2 = 2 - 12x^{-1} + 9x^{-2}.$$ Then we see that the constant term of $f$ is given by $$1 \cdot 2 + \binom{n}{1} (x/5)(-12x^{-1}) + \binom{n}{2}(x/5)^2(9x^{-2}),$$ since these are the only terms for which the power of $x$ will be zero.
As usual, the binomial expansion helps:
$$
\left(2x - \frac 1x\right)^n = \sum_{k=0}^n (-1)^k\binom nk\frac{1}{x^k}(2x)^{n-k} = \sum_{k=0}^n \binom nk (-1)^k2^{n-k}x^{n-2k}
$$
Now, when does the constant term appear? Precisely when $n=2k$, as one sees above. Also, if $n$ is odd, then you see that the constant term is zero, so $n$ must be even.
Then, the constant itself is $(-1)^k\binom nk 2^{n -k } = \binom n{\frac n2} 2^{\frac n2}$.
Now, this is equal to $-160$, as you say. We note that $160$ is a multiple of $5$, hence $n \geq 5$ must happen, otherwise $\binom nk$ cannot be a multiple of $5$. Also, $k$ is odd, as the coefficient is negative, hence $n = 6,10,...$ As we notice, $n=6$ gives $\binom 63 \times 8 = 20 \times 8 = 160$. Hence, the answer is $n = 6$.
Almost as if to prove a point:
$ \left(2x - \frac 1x\right)^6 = 64 x^6 + \frac 1{x^6} - 192 x^4 - \frac{12}{x^4} + 240 x^2 + \frac{60}{x^2} - 160$.
Best Answer
Hint: equalize the two expressions that you have already found for the $5^{th}$ and $6^{th}$ terms, and solve the resulting equation for $\frac{a}{b}$. In doing this, write the binomial numbers using factorials.