= Solution
The strain is an <Ornstein-Uhlenbeck process>. Multiplying its equation by the <integrating factor> $e^{t/\tau}$ and using the zero initial condition gives
$$
\widetilde\sigma(t)=\int_0^t e^{-(t-s)/\tau}f(s)\,ds=\sqrt\kappa\int_0^t e^{-(t-s)/\tau}\,dW_s.
$$
The second expression is a <stochastic integral> against <Brownian motion>. A deterministic linear functional of <Gaussian white noise> is <Gaussian>, and its <expected value> is zero. For $t\ge t'$, the noise <covariance> gives
$$
\begin{aligned}
\mathbb E[\widetilde\sigma(t)\widetilde\sigma(t')]
&=\kappa\int_0^{t'}e^{-(t-s)/\tau}e^{-(t'-s)/\tau}\,ds\\
&=\frac{\kappa\tau}{2}\left[e^{-(t-t')/\tau}-e^{-(t+t')/\tau}\right].
\end{aligned}
$$
For $t'\gg\tau$, the initial-condition term is negligible. The stationary <autocorrelation> is therefore
$$
\boxed{\mathbb E[\widetilde\sigma(t)\widetilde\sigma(t')]\simeq\frac{\kappa\tau}{2}e^{-|t-t'|/\tau}.}
$$
Its normalized exponential decay has <correlation time> $\tau$. The <zero-start Ornstein-Uhlenbeck covariance> also shows how stationarity is approached.
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