Find the determinant of $A$ satisfying $A^{-1}=I-2A.$

No such $A$ exists. Hence we cannot speak of its determinant.

Suppose $A$ is real and $A^{-1}=I-2A$. Then $A^2-\frac12A+\frac12I=0$. Hence the minimal polynomial $m_A$ of $A$ must divide $x^2-\frac12x+\frac12$, which has no real root. Therefore $m_A(x)=x^2-\frac12x+\frac12$. But the minimal polynomial and characteristic polynomial $p_A$ of $A$ must have identical irreducible factors, and this cannot happen because $p_A$ has degree 3 and $m_A$ is an irreducible polynomial of degree 2.

Edit: The OP says that the question appears on an extrance exam paper, and four answers are given: (a) $1/2$, (b) $−1/2$, (c) $1$, (d) $2$. It seems that there's a typo in the exam question and $A$ is probably 2x2. If this is really the case, then the above argument shows that the characteristic polynomial of $A$ is $x^2-\frac12x+\frac12$. Hence $\det A = 1/2$.

Let $\lambda$ be a real eigenvalue with eigenvector $x$ (there is a real root to the characteristic equation). Since $A$ is invertible, $\lambda\neq 0$, so $Ax = \lambda x$ and $A^{-1}x=\lambda^{-1}x.$ Putting these into $A^{-1}x=x-2Ax$ gives $2\lambda^2-\lambda+1=0,$ contradicting that $\lambda$ is real. Hence no such $A$ exists.

It is now abundantly clear there was a typo in the question. I showed above no such real matrix exists. Even if we allow complex entries, the characteristic polynomial has the form $$(x^2-x/2+1/2)(x-z)$$ for some $z\in \mathbb{C}\setminus{\mathbb{R}}$ and the determinant is not real, so not one of the options.