The force required to punch a square hole 2cm side in steel sheet 2mm thick is:
(shearing stress of steel sheet $=3.5\times {{10}^{8}}N/{{m}^{2}}$)
$\begin{align}
  & \left( 1 \right)5.6\times {{10}^{4}}N \\
 & \left( 2 \right)3.4\times {{10}^{4}}N \\
 & \left( 3 \right)9.1\times {{10}^{4}}N \\
 & \left( 4 \right)6.8\times {{10}^{4}}N \\
\end{align}$

Hint: To answer this question we will use the concept of stress and strain. Obtain the relation between the shearing stress and the surface area and force. Find the surface area of the boundary of the hole. Then put the given values on the obtained expression to find the required answer.

Complete answer:
A force is exerted on a steel sheet to punch a square hole. The shearing stress is exerted on the rectangular surface of the hole i.e. the boundary of the hole.
Now, the side of the square hole is $a=2cm=0.02m$
The thickness of the sheet is $b=2mm=0.002m$
So, the boundary surface area of the hole is $A=4ab=4\times 0.02\times 0.002{{m}^{2}}=1.6\times {{10}^{-4}}{{m}^{2}}$
Now, shearing stress can be defined as a stress that acts in the plane of the cross section of the material. The stress on a material due to an applied force can be defined as the force per unit area of the surface of the material.
The minimum shearing stress to punch a hole in the steel can be expressed as,
Shearing stress $=\dfrac{F}{A}$
$\begin{align}
  & \dfrac{F}{1.6\times {{10}^{-4}}}=3.5\times {{10}^{8}} \\
 & F=3.5\times {{10}^{8}}\times 1.6\times {{10}^{-4}} \\
 & F=5.6\times {{10}^{4}}N \\
\end{align}$
So, the force required to punch a square hole of 2cm in a steel sheet of 2mm thick is $5.6\times {{10}^{4}}N$.

So, the correct answer is “Option 1”.

Note:
The shearing stress on an object can be defined as the force applied tangentially to the surface of the object per unit area of the object. When we punch a hole in the steel sheet, we will apply a force on the boundary of the hole and this is a shearing stress.