The evidence is the <convolution of independent random variables>[convolution] of $N(0,\tau^2)$ and $N(0,\sigma^2)$, so
$$
Z=p(y)=\frac1{\sqrt{2\pi(\sigma^2+\tau^2)}}
\exp\!\left[-\frac{y^2}{2(\sigma^2+\tau^2)}\right].
$$
Part i now gives the fully simplified expectation
$$
\mathbb E_{\theta\mid y}\widehat I
=\sqrt{2\pi(\sigma^2+\tau^2)}
\exp\!\left[\frac{y^2}{2(\sigma^2+\tau^2)}\right].
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 219 / 4 / a / iii / Solution

If independent real random variables $X$ and $Y$ have densities $f_X$ and $f_Y$, their sum has the <convolution> density
$$
f_{X+Y}(z)=(f_X*f_Y)(z)=\int_{-\infty}^{\infty}f_X(x)f_Y(z-x)\,dx.
$$

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 Incoming links: Convolution of independent random variables