Answer
$${\text{sixth term}} = \frac{{11!}}{{5!6!}}\left( { - 32{m^{ - 5}}} \right)\left( {729{n^{ - 12}}} \right)$$
Work Step by Step
$$\eqalign{
& {\text{The middle term of }}{\left( { - 2{m^{ - 1}} + 3{n^{ - 2}}} \right)^{11}} \cr
& {\text{The binomial has 12 terms, then the middle terms are: the sixth }} \cr
& {\text{and the seventh term}} \cr
& \cr
& {\text{In the sixth term, }}3{n^{ - 2}}{\text{ has an exponent of 6}} - 1,{\text{ or 5, }} \cr
& {\text{while }} - 2{m^{ - 1}}{\text{ has an exponent of 11}} - 5,{\text{ or 6}}{\text{.}} \cr
& {\text{Using the }}k{\text{th Term of The Binomial Expansion}} \cr
& {\text{sixth term}} = \left( {11{\bf{C}}5} \right){\left( { - 2{m^{ - 1}}} \right)^6}{\left( {3{n^{ - 2}}} \right)^5} \cr
& {\text{sixth term}} = \frac{{11!}}{{6!5!}}{\left( { - 2{m^{ - 1}}} \right)^6}{\left( {3{n^{ - 2}}} \right)^5} \cr
& {\text{sixth term}} = 462\left( {64{m^{ - 6}}} \right)\left( {243{n^{ - 10}}} \right) \cr
& {\text{sixth term}} = 7185024{m^{ - 6}}{n^{ - 10}} \cr
& \cr
& {\text{In the seventh term, }}3{n^{ - 2}}{\text{ has an exponent of 7}} - 1,{\text{ or 6, }} \cr
& {\text{while }} - 2{m^{ - 1}}{\text{ has an exponent of 11}} - 6,{\text{ or 5}}{\text{.}} \cr
& {\text{Using the }}k{\text{th Term of The Binomial Expansion}} \cr
& {\text{sixth term}} = \left( {11{\bf{C}}6} \right){\left( { - 2{m^{ - 1}}} \right)^5}{\left( {3{n^{ - 2}}} \right)^6} \cr
& {\text{sixth term}} = \frac{{11!}}{{5!6!}}\left( { - 32{m^{ - 5}}} \right)\left( {729{n^{ - 12}}} \right) \cr
& {\text{sixth term}} = 462\left( { - 32{m^{ - 5}}} \right)\left( {729{n^{ - 12}}} \right) \cr
& {\text{sixth term}} = - 10777536{m^{ - 5}}{n^{ - 12}} \cr} $$
