$(2a-1)^n$ has the form $ \underbrace{(2a-1)(2a-1)(2a-1)\cdots(2a-1)}_{n-\rm terms} $
To obtain the product, you would keep applying the distributive law, again and again, until there were no multiplications to perform. For instance $\color{darkgreen}{(2a-1)}\color{darkblue}{(2a-1)}=2a\color{darkblue}{(2a-1)}-1 \color{darkblue}{(2a-1)}=\color{maroon}{2a\cdot2a-2a-2a+1}=4a^2-4a+1.$
Please note that the maroon expression above could have been obtained by summing all products obtained from selecting a term (either $2a$ or $-1$) of the green expression on the left and a term of the blue expression on the left to form the product (this is why we have the so-called "FOIL" method; the mnemonic gives all possible products, here).
Going back to $ \tag{1}(2a-1)(2a-1)(2a-1)\cdots(2a-1), $ all these multiplications arising from all that distributing would give in the end (after collecting like terms) an expression of the form: $ a_na^n+a_{n-1}a^{n-1}+\cdots +a_1a+a_0 $ where the $a_i$ are particular numbers.
As in the case above, one can find $(2a-1)^n$, by taking the sum of all possible products obtained by selecting either $2a$ or $-1$ from each of the $n$ factors of $(1)$ to form the product.
Now, how would one get the $a_{n-1}a^{n-1}$ term?
Well, to obtain "part" of it you'd pick a "$\color{darkgreen}{-1}$" from one of the terms in $(1)$ and $\color{maroon}{2a}$ from the others. For example: pick $ (\color{maroon}{2a}{-1})(\color{maroon}{2a}-1)({2a}\color{darkgreen}{-1})\cdots(\color{maroon}{2a}-1), $ and then take the product: $(2a)^{n-1}( -1) $.
But, there are $n $ ways of selecting the $-1$ and each of these gives a $(2a)^{n-1}(-1) $ term. Adding all these together gives the $a_{n-1}a^{n-1}$ term, so: $ a_{n-1}a^{n-1}=n(2a)^{n-1}(-1) =-n2^{n-1}a^{n-1}. $
So $a_{n-1}=-n 2^{n-1}$.
Finally set $-n 2^{n-1}=-192$ and solve for $n$.