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If $tilde{lambda}$ is an eigenvalue of $A$ of multiplicity $n$, then $A$ is a scalar matrix, where $A$ is $n times n$ real symmetric matrix.

So I know that $A = begin{bmatrix}a&b\b&cend{bmatrix} $
and that $$det(A-lambda I_2) = (a-λ)(c-λ) - b^2$$
However I don't really know how to proceed from there.

Hint:
If $A$ has an eigenvalue of multiplicity $n$, it is the only eigenvalue of $A$, so. Furthermore real symmetric matrices are diagonalisable, so its diagonal form is $lambda I$. Can you deduce that $A$ is $lambda I$?.

More details:
Since is real symmetric, there exists an orthogonal matrix $P$ and a diagonal matrix $D$ such that
$$A=PDP^{-1}=P(lambda I)P^{-1}=lambda(PIP^{-1})=lambda I.$$

Your definition of scalar matrix is a diagonal matrix which every element in the diagonal are the same, right?

I assumed that the question should be "... then $A$ is a matrix similar to a scalar matrix". But you also can abuse notation and say that, in fact, $A$ "is'' a scalar matrix by the fact that similiar matrices can represent the same linear transformation...

But, if $A$ is a $ntimes n$ symmetric matrix, then $A$ is diagonalizable and there is a invertible $P$ such that $A=P DP^{-1}$, where $D=diag(lambda,cdots,lambda)$.