Let $f$ be holomorphic on $\overline{D_2}$ with $|f(z)|\leq 1$ for all $z\in\overline{D_2}$.
Show that $|f''(z)|\leq 4$ for all $z\in\overline{D_1}$.
My attempt: By the standard inequality and the Cauchy integrla formula we have for all $z\in\overline{D_1}$: $$|f''(z)|=\left|\frac{2!}{2\pi i}\oint_{\partial D_2}\frac{f(w)}{(w-z)^3}dw\right|=\frac{1}{\pi }\left|\oint_{\partial D_2}\frac{f(w)}{(w-z)^3}dw\right|\leq$$ $$\leq\frac{1}{\pi}\cdot 2\pi \max_{|w|=2}\left|\frac{f(w)}{(w-z)^3}\right|=2\max_{|w|=2}\frac{|f(w)|}{|w-z|^3}$$ Now I want to use $|f(w)|\leq 1$ since $D_1\subset D_2$ and also the triangle inequality: $$2\max_{|w|=2}\frac{|f(w)|}{|w-z|^3}\leq\max_{|w|=2} \frac{2}{||w|-|z||^3}=\frac{2}{|2-|z||^3}$$ And since $|z|\leq 1$ this is always less or equal to $2$. So in total $|f''(z)|\leq 2$ on the given disk which is obviously less than $4$. Is this correct?
Your method is correct but the length of the circle of radius $2$ is $4\pi$. You took it as $2\pi$ in your first inequality. So the final bound is indeed be $4$!