Answer
$$y = \frac{{240}}{{1 + 15{e^{ - \left( {3/20} \right)t}}}}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dt}} = \frac{{3y}}{{20}} - \frac{{{y^2}}}{{1600}},{\text{ }}\left( {0,15} \right) \cr
& {\text{Factoring}} \cr
& \frac{{dy}}{{dt}} = \frac{3}{{20}}y\left( {1 - \frac{y}{{240}}} \right) \cr
& {\text{The differential equation is in the form }}\frac{{dy}}{{dt}} = ky\left( {1 - \frac{y}{L}} \right) \cr
& \underbrace {\frac{{dy}}{{dt}} = \frac{3}{{20}}y\left( {1 - \frac{y}{{240}}} \right)}_{\frac{{dy}}{{dt}} = ky\left( {1 - \frac{y}{L}} \right)} \to k = \frac{3}{{20}},{\text{ }}L = 240 \cr
& {\text{The general solution is }}y = \frac{L}{{1 + b{e^{ - kt}}}}{\text{ }}\left( {{\text{Example 6, page 419}}} \right) \cr
& {\text{Therefore, substituting }}k = \frac{3}{{20}}{\text{ and }}L = 240 \cr
& \underbrace {y = \frac{L}{{1 + b{e^{ - kt}}}}}_ \downarrow \cr
& y = \frac{{240}}{{1 + b{e^{ - \left( {3/20} \right)t}}}}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Using the initial condition }}\left( {0,15} \right) \cr
& 15 = \frac{{240}}{{1 + b{e^0}}} \cr
& 15\left( {b + 1} \right) = 240 \cr
& b + 1 = 16 \cr
& b = 15 \cr
& {\text{Substituting }}b{\text{ into }}\left( {\bf{1}} \right) \cr
& y = \frac{{240}}{{1 + 15{e^{ - \left( {3/20} \right)t}}}} \cr} $$
