This is in some sense a follow-up to this question.

The answer there says that over $\mathsf{Z}_2$ (second-order arithmetic), (boldface) $\mathbf{\Sigma}^1_1$-determinacy is enough to entail the existence of $0^{\sharp}$, which, if I'm understanding correctly, gives you the existence of a countable transitive model of $\mathsf{ZFC}$ (over just $\mathsf{Z}_2$).

Since the existence of a countable transitive model of $\mathsf{ZFC}$ is a single sentence, we know by compactness that only some fragment of $\mathsf{Z}_2 + \mathbf{\Sigma}^1_1\text{-Det}$ is necessary to entail the existence of a countable transitive model of $\mathsf{ZFC}$, so I'm wondering whether one of the standard subsystems of second-order arithmetic is actually sufficient.

It's a classic result of Steel that open determinacy is equivalent to $\mathsf{ATR}_0$ over $\mathsf{RCA}_0$, so if we restrict attention to the big five, there's only really two options.

Question.Does $\mathsf{RCA}_0 + \mathbf{\Sigma}^1_1\text{-Det}$ (equivalently $\mathsf{ATR}_0 + \mathbf{\Sigma}^1_1\text{-Det}$) entail the existence of a countable transitive model of $\mathsf{ZFC}$? If not does $\mathbf{\Pi}^1_1\text{-}\mathsf{CA}_0+\mathbf{\Sigma}^1_1\text{-Det}$ suffice?

I'm also curious about analogous questions regarding just consistency strength and possibly only considering (lightface) $\Sigma^1_1$-determinacy.