As $l^2=25 {\mathrm{cm}}^2$, then $l=5 \mathrm{cm}=0.05 \mathrm{m}$.

Given: $t=1 \mathrm{s}$

Velocity of the square loop, $v=\frac{0.05}{1}=0.05 \mathrm{m} {\mathrm{s}}^{-1}$.

Induced current,

$$\begin{gathered} i=\frac{V}{R}=\frac{Blv}{R} \\ =\frac{40\times 0.05\times 0.05}{10}=0.01 \mathrm{A} \end{gathered}$$

Now force acting on the side of the square loop, $F=Bil=40\times 0.01\times 0.05$

$\Rightarrow F=0.02 \mathrm{N}$

Therefore, work done

$$\begin{gathered} W=Fl \\ =0.02\times l \\ =0.02\times 0.05 \end{gathered}$$

$\Rightarrow W=1\times {10}^{-3} \mathrm{J}$