Writing $X_i=\theta(1+W_i)$ with $W_i\stackrel{iid}\sim U(0,1)$ (since $X_i/\theta-1$ is uniform on $(0,1)$ exactly when $X_i\sim U(\theta,2\theta)$ ), both order statistics satisfy $Y_{(1)}=\theta(1+W_{(1)})>\theta$ and $Y_{(2)}=\theta(1+W_{(2)}) >\theta$ almost surely, since $W_{(1)},W_{(2)}>0$ a.s. — so $\theta<Y_{(1)}\le Y_{(2)}$ always, meaning $\theta$ can never lie inside $[Y_{(1)},Y_{(2)}]$ under this literal reading (giving coverage probability $0$ , matching none of the printed options).

By contrast, the general and standard fact that, for a sample of size $2$ from any continuous distribution with population median $m$ , $P(Y_{(1)}<m<Y_{(2)})=1-P(\text{both above }m)-P(\text{both below }m)=1-\tfrac14-\tfrac14=\tfrac12$ would cleanly give the keyed answer $1/2$ if $\theta$ here denoted the population median of the sampling distribution rather than its lower endpoint. We record the official answer, while flagging that the literal $U(\theta,2\theta)$ reading (where $\theta$ is the lower endpoint, not the median) gives coverage probability $0$ instead.