Answer
$6\;\rm T$
Work Step by Step
We know that the net magnetic flux through a closed surface is zero.
So,
$$\oint \Phi_m\cdot dA=0$$
In this scenario, we are presented with a total of six surfaces. The first pair consists of the upper surface and the bottom surface, both of which possess equal areas. Similarly, we have the right surface and the left surface, which also share the same area. Finally, we have the front and back surfaces, each with identical areas.
And since the area of the 6 surfaces is constant, so $\Phi_{\rm surface}=BA\cos\theta$.
Thus,
$$ B_{\rm up} A_{\rm up}\cos\theta_{\rm up}+ B_{\rm bot} A_{\rm bot}\cos\theta_{\rm bot}+B_{\rm right} A_{\rm right}\cos\theta_{\rm right}+ B_{\rm left} A_{\rm left}\cos\theta_{\rm left}+ B_{\rm front} A_{\rm front}\cos\theta_{\rm front}+B_{\rm back} A_{\rm back}\cos\theta_{\rm back}=0$$
Plugging from the given figure,
$$ (2)(0.02\times 0.01)\cos180^\circ+
(1)(0.02\times 0.01)\cos180^\circ+
(2)(0.02\times 0.01)\cos0^\circ+
(3)(0.02\times 0.01)\cos0^\circ+
(2)(0.01\times 0.01)\cos0^\circ+
(B_{\rm back} )(0.01\times 0.01)\cos\theta
=0$$
Therefore,
$$B_{\rm back}(0.01\times 0.01)\cos\theta =-0.0006$$
And since it is negative, the angle had to be 180$^\circ$
$$B_{\rm back} =\dfrac{-0.0006}{0.01^2}=-6\;\rm T$$
$$B_{\rm back} =\color{red}{\bf 6}\;\rm T\tag{Into the back surface}$$
