eigenvalue of block matrix in terms of original matrix A is a $4*4$ matrix with eigenvalues $\lambda_A$. Consider a block matrix $B = \left( \begin{array}{ccc}
A & I \\
I & A \end{array} \right) $. Then how can we find eigenvalue $\lambda_B$ of matrix B in terms of $\lambda_A$ ?
I know that to find $\lambda_B$, we have to do $|B-\lambda_BI| = 0$, and by properties of block matrices, we have equivalently $|(A-\lambda_BI)^2 - I^2| = 0$.
But I am stuck at this point, I don't know how to proceed further.

It looks like you are trying to find a determinant of block matrix using the rule for 2x2 matrices, but it is incorrect.
According to http://en.wikipedia.org/wiki/Determinant#Block_matrices 
$$
|B-\lambda_bI| = 
\left|
\left(\begin{matrix}A-\lambda_bI & I \\ I & A - \lambda_bI\end{matrix}\right)
\right|=
|A-\lambda_bI - I||A-\lambda_bI + I|= \\
=|A-(\lambda_b+1)I||A-(\lambda_b - 1)I| = 0
$$
Then you can determine $\lambda_b$ using the fact that $\lambda_a$ is a solution of $|A - \lambda_aI|=0$. 
