Follows the enunciated of the question and after my reasoning. In short, I didn't arrived to the answer of the book but I know the reasoning. Follows the question:
Follows the hint and the answer of the book:
My reasoning:
Let A = \begin{pmatrix}1_{m} & 0\\1_{m}&1_{m}\\1_{n}&21_{n}\end{pmatrix}, then
$A^{t}A = \begin{pmatrix}1_{m} & 1_{m}& 1_{n}\\0&1_{m}&-21_{n}\end{pmatrix} \times \begin{pmatrix}1_{m} & 0 \\1_{m}&1_{m}\\1_{n}&-21_{n}\end{pmatrix}= \begin{pmatrix}2m+n & m-2n \\m-2n&m-4n\end{pmatrix}$
Then
$(A^{t}A)^{-1}=\frac{1}{(2m+n)(m-4n)-(m-2n)^{2}}\begin{pmatrix}m-4n & 2n-m \\2n-m&2m+n\end{pmatrix}$
After:
$\hat{\beta} = (A^{t}A)^{-1}A^{t}Y$
Then, using the inverse matrix calculated above
$\hat{\beta} = \frac{1}{(2m+n)(m-4n)-(m-2n)^{2}}\begin{pmatrix}m-4n & 2n-m \\2n-m&2m+n\end{pmatrix}\begin{pmatrix}1_{m} & 1_{m} & 1_{n} \\0&1_{m}&-21_{n}\end{pmatrix}\begin{pmatrix}U \\V\\W\end{pmatrix}$
I don't getting much advanced when I develop the expression above. I know the way is this, but i need some help to arrive to the answer.
Don't have enough rep to comment. It seems that you are using $AY$ instead of $A^tY$ in your expression for $\hat{\beta}$. Additionally the bottom right entry of $A^tA$ should be $m+4n$.