at the instant t=0 a force F=kt (k is a constant) acts on a small body of mass m resting on a smooth horizontal surface.the time, when body leaves the surface is,
a)mgksinα b)ksinα/mg c)mgsinα/k d)mg/ksinα

From the question what I understand the force is applied at an angle $\alpha$ from the horizontal. The above figure represent the situation. The block will leave the surface when the vertical component of the applied force $\overrightarrow{F}$ will be equal to the blocks weight. Now vertical component of the force is $F\sin \alpha$. So we have
$$\begin{gathered} F\sin \alpha =mg \\ \Rightarrow Kt\sin \alpha =mg \\ \Rightarrow t=\frac{mg}{K\sin \alpha } \end{gathered}$$
d) is the correct answer.