The correct answer is (a) \(\frac{s}{s^2+ω^2}\)
For explanation I would say: Laplace transform, L{x(t)} = X(s) = \(\int_{-∞}^∞ x(t) e^{-st} \,dt\)
X(s) = L{cosωt u(t)} = \(L[\frac{e^{jωt} + e^{-jωt}}{2} \,u(t)] = \frac{1}{2} L[e^{jωt} \,u(t)] + \frac{1}{2} L[e^{jωt} \,u(t)]\)
X(s) = \(\frac{1}{2} (\frac{1}{s-jω}) + \frac{1}{2} (\frac{1}{s+jω}) = \frac{s}{s^2+ω^2}\).