1. Area of the square loop

The side of the square is $a = 10\text{ cm} = 0.10\text{ m}$ So its area is $A = a^2 = (0.10)^2 = 0.01\text{ m}^2$

2. Angle between magnetic field and area vector

The loop is placed in the east-west plane. So the normal to the plane is along the north-south direction.

The magnetic field is in the north-east direction. Hence, the angle between the magnetic field and the normal is $\theta = 45^\circ$

3. Initial magnetic flux

Magnetic flux through the loop is $\Phi = BA\cos\theta$ Thus initially, $\Phi_i = (0.20)(0.01)\cos 45^\circ$ $\Phi_i = 0.002\cdot \frac{1}{\sqrt{2}}$ $\Phi_i = \sqrt{2}\times 10^{-3}\text{ Wb}$

4. Final magnetic flux

The magnetic field decreases to zero in $1\text{ s}$, so final flux is $\Phi_f = 0$

5. Induced emf

Using Faraday's law, $|\mathcal{E}| = \left|\frac{\Delta \Phi}{\Delta t}\right|$ $|\mathcal{E}| = \frac{\Phi_i - \Phi_f}{1} = \sqrt{2}\times 10^{-3}\text{ V}$

Given, $|\mathcal{E}| = \sqrt{x}\times 10^{-3}\text{ V}$ Comparing, $\sqrt{x} = \sqrt{2}$ So, $x=2$

6. Comparison with stored answer

Derived answer: $2$

Stored correct answer: $2$

They match.