Answer
$\frac{\sqrt 3}{30}m/sec$, increasing.
Work Step by Step
Step 1. Point-A is moving along the x-axis; we can assume its coordinates as $(a, 0)$. Point-B is moving along the y-axis; we can assume its coordinates as $(0, b)$.
Step 2. Using similar triangles shown in the figure, we get $\frac{b}{r}=\frac{\sqrt {a^2+b^2}}{a} \longrightarrow
\frac{1}{r^2}=\frac{a^2+b^2}{a^2b^2}=\frac{1}{a^2}+ \frac{1}{b^2}=a^{-2}+b^{-2}$
Step 3. Taking the derivative of the equation above with $r$ as a constant, we get $-2a^{-3}a'-2b^{-3}b'=0$ or $a'=-\frac{a^3}{b^3}b'$
Step 4. With $b=2r, b'=-0.3r$, we have $\frac{2r}{r}=\frac{\sqrt {a^2+(2r)^2}}{a}\longrightarrow 4a^2=a^2+4r^2 \longrightarrow a^2=\frac{4}{3}r^2 $
Step 5. Using the result from step 3 and 4, we have $\frac{da}{dt}=a'=-\frac{a^3}{(2r)^3}(-0.3r)=\frac{3}{80r^2}\frac{4}{3}r^2(\frac{4}{3}r^2)^{1/2}=\frac{\sqrt 3}{30}m/sec$ and this means that $OA$ is increasing.
