= Solution
For arbitrary $\beta$,
$$
\mathbb E\|\widehat\beta\|^2
=\|\beta\|^2
+\operatorname{tr}\!\left(\sigma^2(X^TX)^{-1}\right)
=\|\beta\|^2+\sigma^2\sum_{j=1}^p\lambda_j^{-1}.
$$
Independence from part (ii) and $Z=n\widehat\sigma^2/\sigma^2\sim\chi^2_{n-p}$ give
$$
\mathbb ET
=\frac{\mathbb E\|\widehat\beta\|^2}{\sigma^2}
\mathbb E\!\left(\frac1Z\right).
$$
Because $n-p>2$,
$$
\mathbb E(Z^{-1})=\frac1{n-p-2}.
$$
Thus
$$
\boxed{
\mathbb ET
=\frac{\|\beta\|^2/\sigma^2
+\sum_{j=1}^p\lambda_j^{-1}}
{n-p-2}
}.
$$
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