The answer is $\boxed{\frac{29}{53}}$.
The sequence is a geometric sequence with a common ratio of $\frac{2}{3}$. This means that each term is $\frac{2}{3}$ times the term before it.
The first term is $\frac{4}{9}$. The second term is $\frac{4}{9} \times \frac{2}{3} = \frac{9}{20}$. The third term is $\frac{9}{20} \times \frac{2}{3} = \frac{18}{30}$. The fourth term is $\frac{18}{30} \times \frac{2}{3} = \frac{29}{53}$.
Here is a table of the terms of the sequence:
Term | Value ------- | -------- 1 | $\frac{4}{9}$ 2 | $\frac{9}{20}$ 3 | $\frac{18}{30}$ 4 | $\frac{29}{53}$
Here is a graph of the sequence:
[asy] unitsize(1 cm);
real ticklen = 1; real tickspace = 1; real axisarrowsize = 0.14inch; real tickdown = -0.5; real tickdownlength = -0.12inch; real tickdownbase = -0.25inch; real wholetickdown = tickdown; real wholetickdownlength = tickdownlength; real wholetickdownbase = tickdownbase; real axiswidth = 0.12inch; real linetype = 1; real linewidth = 0.7pt; real dotsep = 3pt; real dotsize = 3pt;
real g = 1.2;
real ticklength = 0.14inch; real axismargin = 0.1inch; real xmargin = 1.2inch; real ymargin = 1.2inch;
draw((0,-1.2)--(0,1.2),Arrows(0.14inch)); draw((-1.2,0)--(1.2,0),Arrows(0.14inch));
label("$x$",(1.2,0),SE); label("$y$",(0,1.2),NW);
real i; for (i=-1; i<=1; ++i) { draw((i,-0.1)--(i,0.1)); }
real j; for (j=-1; j<=1; ++j) { draw((0,j)--(0.1,j)); }
real t; for (t=0; t<=1; t+=0.1) { draw((t,-0.05)--(t,0.05)); }
real x; real y;
x = 0; y = 4/9; draw((x,y)--(x+0.1,y),linetype,linewidth);
x = 0; y = 9/20; draw((x,y)--(x+0.1,y),linetype,linewidth);
x = 0; y = 18/30; draw((x,y)--(x+0.1,y),linetype,linewidth);
x = 0; y = 29/53; draw((x,y)--(x+0.1,y),linetype,linewidth);
label("$\frac{4}{9}$",(-0.1,4/9),SE); label("$\frac{9}{20}$",(-0.1,9/20),SE); label("$\frac{18}{30}$",(-0.1,18/30),SE); label("$\frac{29}{53}$",(-0.1,29/53),SE); [/asy]
