Answer
$${\text{fifteenth term}} = 9743085{a^6}{b^{11}}$$
Work Step by Step
$$\eqalign{
& {\text{tenth term of }}{\left( {{a^3} + 3b} \right)^{11}} \cr
& {\text{In the tenth term, }}3b{\text{ has an exponent of 10}} - 1,{\text{ or 9, }} \cr
& {\text{while }}x{\text{ has an exponent of 11}} - 9,{\text{ or 2}}{\text{.}} \cr
& \cr
& {\text{Using the }}k{\text{th Term of The Binomial Expansion}} \cr
& {\text{fifteenth term}} = \left( {11{\bf{C}}9} \right){\left( {{a^3}} \right)^2}{\left( {3b} \right)^{11}} \cr
& {\text{fifteenth term}} = \frac{{11!}}{{2!9!}}\left( {{a^6}} \right)\left( {177147{b^{11}}} \right) \cr
& {\text{fifteenth term}} = 55\left( {{a^6}} \right)\left( {177147{b^{11}}} \right) \cr
& {\text{fifteenth term}} = 9743085{a^6}{b^{11}} \cr} $$
