= Solution
Because $x_*\sim N_p(0,I)$ is independent, conditional prediction risk equals $\mathbb E(\lVert\widehat\beta_\lambda-\beta^0\rVert_2^2\mid X)$. With $\widehat\Sigma=X^TX/n$ and $\lambda=n\ell$,
$$
\widehat\beta_\lambda-\beta^0
=-\ell(\widehat\Sigma+\ell I)^{-1}\beta^0
+\frac1n(\widehat\Sigma+\ell I)^{-1}X^T\varepsilon.
$$
The noise term has conditional mean zero and covariance $\frac{\sigma^2}{n}\,(\widehat\Sigma+\ell I)^{-1}\widehat\Sigma(\widehat\Sigma+\ell I)^{-1}$. Taking squared norms proves
$$
R_X(\widehat\beta_\lambda)=
\ell^2(\beta^0)^T(\widehat\Sigma+\ell I)^{-2}\beta^0
+\frac{\sigma^2}{n}\operatorname{tr}\bigl(\widehat\Sigma(\widehat\Sigma+\ell I)^{-2}\bigr).
$$
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