It has been given that the coefficient of static friction between the box and the train's floor is $0.2$. The maximum acceleration of the train in which the box lying on its floor will remain at rest is given as,
(let us assume that the value of acceleration due to gravity is $g=10m{{s}^{-2}}$)
$\begin{align}
  & A.2m{{s}^{-2}} \\
 & B.4m{{s}^{-2}} \\
 & C.6m{{s}^{-2}} \\
 & D.8m{{s}^{-2}} \\
\end{align}$

Hint: The value of the static frictional force is given as the product of the coefficient of static friction and the normal force acting over the body. The body will be remaining at the rest itself as long as the force we apply is less than static frictional force. Therefore the maximum value of force applied up to which body is at rest will be the value of static frictional force. This information will help you in solving this question.

Complete step-by-step answer:
First of all the acceleration of the train up to which the static friction acts can be found using the equation,
$ma={{f}_{s}}\le {{\mu }_{s}}N={{\mu }_{s}}mg$
Where $m$ be the mass of the body, $a$ be the acceleration of the train, ${{f}_{s}}$ be the static frictional force, ${{\mu }_{s}}$ be coefficient of static friction, $N$ be the normal force acting on the body, $g$ be the acceleration due to gravity.
It has been mentioned in the question that the coefficient of static friction is given as,
${{\mu }_{s}}=0.2$
The body will be remaining at the rest itself as long as the force we apply is less than static frictional force.
$a\le {{\mu }_{s}}g$
Therefore the maximum value of force applied up to which body is at rest will be the value of static frictional force.
Therefore we can write that,
${{a}_{\max }}={{\mu }_{s}}g$
Substituting the values in it will give,
${{a}_{\max }}=0.2\times 10=2m{{s}^{-2}}$
Therefore the correct answer is mentioned as option A.