= Solution
Combining the Gaussian likelihood with the prior $\theta\sim N_p(0,c^2I_p)$ gives the posterior precision and mean
$$
I_p+c^{-2}I_p,
\qquad
\left(I_p+c^{-2}I_p\right)^{-1}X.
$$
Thus the posterior in the <Gaussian normal-mean shrinkage Bayes estimator> is
$$
\theta\mid X
\sim N_p\left(
\frac{c^2}{1+c^2}X,
\frac{c^2}{1+c^2}I_p
\right).
$$
The <Bayes estimator under squared error loss> is the posterior mean, so
$$
\boxed{\widehat\theta_c(X)=\frac{c^2}{1+c^2}X.}
$$
Its posterior expected loss is the trace of the posterior covariance and does not depend on $X$. Therefore its <Bayes risk> is
$$
\boxed{r(\pi_c)=\frac{pc^2}{1+c^2}.}
$$
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