Compute $\cos \left( \arcsin \frac{2}{3} \right).$
Consider a right triangle where the opposite side is 2 and the hypotenuse is 3.

[asy]
unitsize (1 cm);

draw((0,0)--(sqrt(5),0)--(sqrt(5),2)--cycle);

label("$\sqrt{5}$", (sqrt(5)/2,0), S);
label("$3$", (sqrt(5)/2,1), NW);
label("$2$", (sqrt(5),1), E);
label("$\theta$", (0.7,0.3));
[/asy]

Then $\sin \theta = \frac{2}{3},$ so $\theta = \arcsin \frac{2}{3}.$  By Pythagoras, the adjacent side is $\sqrt{5},$ so $\cos \theta = \boxed{\frac{\sqrt{5}}{3}}.$
Consider a right triangle where the opposite side is 2 and the hypotenuse is 3.

[asy]
unitsize (1 cm);

draw((0,0)--(sqrt(5),0)--(sqrt(5),2)--cycle);

label("$\sqrt{5}$", (sqrt(5)/2,0), S);
label("$3$", (sqrt(5)/2,1), NW);
label("$2$", (sqrt(5),1), E);
label("$\theta$", (0.7,0.3));
[/asy]

Then $\sin \theta = \frac{2}{3},$ so $\theta = \arcsin \frac{2}{3}.$  By Pythagoras, the adjacent side is $\sqrt{5},$ so $\cos \theta = \boxed{\frac{\sqrt{5}}{3}}.$
The final answer is $$\frac{\sqrt{5}}{3}$$. I hope it is correct.