How to find volume from Washer method?

From your graph it seems that you are considering the solid obtained by rotating the curve about the $y$-axis. Is it correct?

Why don't you simplify before integrating? $$(\sqrt{4y})^2-(\sqrt{4y}-0.1)^2=0.4 \sqrt{y}-0.01.$$ It remains to evaluate $$\pi\left(0.4 \int_0^2 \sqrt{y}\,dy -0.01 \int_0^2 \,dy\right)= \pi\left(0.4 \left[\frac{2y^{3/2}}{3}\right]_0^2 -0.01 \cdot 2\right).$$ What is the final result?

If you're rotating about the $x$-axis, you need functions in terms of $x$ for the first quadrant:

$$ x=\sqrt{4y}\implies y=\frac{x^2}{4},\\ x=\sqrt{4y}-0.1\implies y=\frac{(x+0.1)^2}{4}. $$

Your integral should look like this where $2$ is supposedly the upper bound of integration:

$$ V=\pi\int_{0}^{2}\left[\left(\frac{x^2}{4}\right)^2-\left(\frac{[x+0.1]^2}{4}\right)^2\right]\,dx. $$

If you're rotating about the $y$-axis, then your integral looks fine. There must have been a mistake in your calculations:

$$ V=\pi\int_{0}^{2}\left[\left(\sqrt{4}y\right)^2-\left(\sqrt{4y}-0.1\right)^2\right]\,dy=\\ \pi\int_{0}^{2}\left[4y-\left(4y-0.2\sqrt{4y}+0.01\right)\right]\,dy=\\ \pi\int_{0}^{2}\left(4y-4y+0.2\cdot 2\sqrt{y}-0.01\right)\,dy=\\ \pi\int_{0}^{2}\left(0.4\sqrt{y}-0.01\right)\,dy=\\ 0.4\pi\int_{0}^{2}\sqrt{y}\,dy-0.01\pi\int_{0}^{2}\,dy=\\ 0.4\pi\left(\frac{2\sqrt{2^3}}{3}-\frac{2\sqrt{0^3}}{3}\right)-0.01\pi(2-0)=\\ 0.8\pi\frac{\sqrt{2^2\cdot 2}}{3}-0.02\pi=\\ 0.8\pi\frac{2\sqrt{2}}{3}-\pi\frac{0.02\cdot 3}{3}=\\ \pi\left(\frac{0.8\cdot 2\sqrt{2}}{3}-\frac{0.06}{3}\right)=\\ \pi\frac{1.6\sqrt{2}-0.06}{3}\approx 2.31\ \text{cubic units}. $$

This is easy to integrate. Just remember that $\int\sqrt{x}\,dx=\frac{2\sqrt{x^3}}{3}+C$ and $\int\,dx=x+C$.