A force of F - 0.5 ~N is applied on lower block as shown in figure....

Maximum acceleration of 1 kg block may be $a_{\max }=\mu g=1 \mathrm{ m} / \mathrm{s}^2$ Common acceleration without relative motion between two blocks may be, $a=\frac{0.5}{3} \mathrm{ m} / \mathrm{s}^2$ Since, $a < a_{\max }$ There will be no relative motion and blocks will move with acceleration $\frac{0.5}{3} \mathrm{ m} / \mathrm{s}^2$. Force of friction by lower block on upper block, \begin{aligned} f & =m a=(1)\left(\frac{0.5}{3}\right)=\frac{1}{6} \mathrm{ N}(\text{ towards \}) \\ \therefore \quad W & =f \times s \\ & =\frac{1}{6} \times 3=0.5 \mathrm{ J} \end{aligned}