Rewrite $\sin \left(2 \sin ^{-1} \frac{w}{3}\right)$ as an algebraic expression in $w$
\[
\sin \left(2 \sin ^{-1} \frac{w}{3}\right)=
\]

Step 1 ：Let's start with the given expression: \(\sin \left(2 \sin ^{-1} \frac{w}{3}\right)\)

Step 2 ：We can use the double angle formula for sine, which is \(\sin(2x) = 2\sin(x)\cos(x)\). Here, \(x = \sin^{-1}\frac{w}{3}\), so we need to express \(\cos(x)\) in terms of \(w\) as well.

Step 3 ：We can use the Pythagorean identity \(\cos^2(x) = 1 - \sin^2(x)\), and since \(\sin(x) = \frac{w}{3}\), we have \(\cos^2(x) = 1 - \left(\frac{w}{3}\right)^2\).

Step 4 ：Taking the square root of both sides, we get \(\cos(x) = \sqrt{1 - \frac{w^2}{9}}\).

Step 5 ：Substituting \(\sin(x) = \frac{w}{3}\) and \(\cos(x) = \sqrt{1 - \frac{w^2}{9}}\) into the double angle formula, we get \(\sin(2x) = 2\cdot\frac{w}{3}\cdot\sqrt{1 - \frac{w^2}{9}}\).

Step 6 ：Simplifying the expression, we get \(\sin(2x) = \frac{2w\sqrt{1 - \frac{w^2}{9}}}{3}\).

Step 7 ：So, the algebraic expression for \(\sin \left(2 \sin ^{-1} \frac{w}{3}\right)\) in terms of \(w\) is \(\boxed{\frac{2w\sqrt{1 - \frac{w^2}{9}}}{3}}\).