Let
$$
\widehat\kappa=\sup_{n<\omega}j^n(\kappa)
$$
be the supremum of the <Kunen critical sequence>, and put
$$
X=j^{\prime\prime}\widehat\kappa
=\{j(\xi):\xi<\widehat\kappa\}.
$$
The required choices are
$$
\boxed{\beta=\widehat\kappa+2,\qquad
\gamma=\widehat\kappa+1,\qquad
X=j^{\prime\prime}\widehat\kappa.}
$$
Indeed $X$ has rank $\widehat\kappa$, so $X\in V_{\widehat\kappa+1}$, while the <Kunen lemma> gives $X\notin M$ once the domain contains the omega-Jonsson function required by the proof, which is ensured by $\alpha\geq\widehat\kappa+2$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 116 / 2 / c / Solution

{c}
{title2=$\langle j^n(\kappa):n<\omega\rangle$}

The critical sequence of $j$ starts with its critical point $\kappa$ and iterates $\kappa_{n+1}=j(\kappa_n)$. Its supremum $\widehat\kappa=\sup_{n<\omega}j^n(\kappa)$ is the least fixed point of $j$ above $\kappa$ whenever the relevant iterates lie in the domain.

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 Incoming links: Kunen critical sequence