Verified Answer :

It would be
2x3xnxn
N has a 1 in front of it it's just not shown
So
2x3x1nx1n then you'd at as much as possible
5x2n

\(2x^3*n*n=54\) 

\(\ Use \ Product \ Rule\)       \(x^ax^b=x^a^+^b\) 

\(2x^3n^2=54\) 

\(x^3n^2= \dfrac{54}{2}\)      \(\ Divide \ both \ sides \ by \ 2\) 

\(x^3n^2=27\) 

\(n^2= \dfrac{27}{x^3}\)        \(\ Divide \ both \ sides \ by \ x^3\) 

\(n=+ \sqrt{ \dfrac{27}{x^3}}\) \(\ Take \ the \ square \ root \ of \ both \ sides\) 

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\(If \ you \ mean \ 2x*3xnxn=54 . \ Lets \ Solve \ for \ n\) 

\(2x*3xnxn=54\) 

\(6x^3n^2=54\) 

\(x^3n^2= \dfrac{54}{6}\)        \(\ Divide \ both \ sides \ by \ 6\)

\(x^3n^2=9\) 

\(n^2= \dfrac{9}{x^3}\)           \(\ Divide \ both \ sides \ by \ x^3\)

\(n=+ \sqrt{ \dfrac{9}{x^3} }\)  \(\ Take \ the \ square \ root \ of \ both \ sides\)