Rule for the ROC of a sum. Each exponential term contributes its own region of convergence, and the ROC of the whole signal is the intersection of them. Two facts fix the shape:

A right-sided term contributes a right half plane bounded by its pole; a left-sided term contributes a left half plane bounded by its pole; and the ROC can never contain a pole.

Apply to the two terms. The first term \(e^{-at}u(t)\) has a pole at \(s=-a\) and converges to its right:

\(\operatorname{Re}(s) \gt -a\)

The second term contributes the pole at \(s=-b\), and taken as the oppositely bounded (left-sided) part of the signal it converges to its left:

\(\operatorname{Re}(s) \lt -b\)

Intersect them. The common region is the vertical strip lying between the two poles:

\(-a \lt R \lt -b\)

which is the keyed answer.

Why the other options cannot be right. The boundaries of an ROC are always the pole locations, and the poles of these exponentials sit at −a and −b, not at +a or +b. So any option written in terms of a and b with positive signs (options 1 and 4, and the mixed option 3) misplaces the boundaries.

The bigger picture. The three possible ROCs for a two-pole X(s) with poles at −a and −b correspond to three different time signals: the right half plane to the right of both poles ⇒ a purely causal signal; the left half plane to the left of both ⇒ a purely anticausal signal; the strip in between ⇒ a two-sided signal. This is why an inverse Laplace transform is unique only when the ROC is specified. A useful corollary: a causal system is stable exactly when its ROC — and therefore the whole right half plane beyond its rightmost pole — includes the jω axis.

Hence, the region of convergence is \(-a \lt R \lt -b\).