If \begin{equation*} A= \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} \end{equation*} put $\alpha={[|a|^2+|b|^2+|c|^2+|d|^2]}^\frac{1}{2}$ and show that $\|A\|=\frac{1}{2}(\alpha^2+\sqrt{\alpha^4-4\delta^2})$, where $\delta^2=det A^*A$.
This is an exercise which related to operators on Hilbert space.I guesses the norm can calculate by Schur test, but I don't know how I use it.
The operator Euclidean norm of $A$ is just the square root of the largest eigenvalue of $A^* A$. But you know the formula for the characteristic polynomial of a 2x2 matrix is
$$p_A(\lambda)=\lambda^2-T(A) \lambda + \Delta(A)$$
where $T$ denotes the trace and $\Delta$ denotes the determinant. Plug in $A^* A$ into this equation and pick the larger solution. Then write out $T(A^*A)$ in terms of $a,b,c,d$ to finish. You could also have written out $\Delta(A^* A)$ in terms of $a,b,c,d$ but that's not necessary because of how the problem is formulated.
(You also have a typo: you have written $\| A \|^2$. This sort of had to be the case just to ensure the units make sense: the left side is linear in the entries of $A$ while the right side is quadratic in the entries of $A$.)