Each of the three conditions the question lists is on its own sufficient to declare the system unstable — option 4 — and it is worth seeing that they are three descriptions of one fact.

How the three connect. A pole at \(s=\sigma+j\omega\) contributes a term

\(e^{\sigma t}\left(A\cos\omega t+B\sin\omega t\right)\)

to the impulse response. If \(\sigma\gt0\) the exponential envelope grows, so g(t) is unbounded — the second statement. And a growing function certainly cannot have a finite integral of its magnitude — the third. Each follows from the first.

Why absolute integrability is the formal criterion. A system is BIBO stable if and only if

\(\int_{0}^{\infty}\left|g(\tau)\right|d\tau\lt\infty\)

The reason is that the output is the convolution of input and impulse response, so a bounded input of magnitude M gives

\(\left|y(t)\right|\le M\int_{0}^{\infty}\left|g(\tau)\right|d\tau\)

A finite integral therefore guarantees a bounded output; an infinite one means some bounded input will drive the output to infinity.

Distinguishing the other three terms, since the question offers them.

Absolutely stable means every pole lies strictly in the left half plane, so the response decays for all values of gain in the range considered.
Marginally stable means a simple pole pair sits exactly on the imaginary axis: the response neither grows nor decays but oscillates for ever at that frequency. Note that repeated poles on the axis give a response growing as \(t\sin\omega t\), which is genuinely unstable.
Critically stable is used for the same boundary condition, the point at which a gain increase would push the poles across into the right half plane.

None of these describes a pole with a positive real part, which lies unambiguously outside the stable region. In a physical system the growth is eventually stopped by saturation or by mechanical failure, but the linear model correctly predicts an unbounded response.

Hence, the system is unstable.