A mass of M kg  is suspended by a weight less string the horizontal force that requires to displace it until the string making an angle 45° with the initial vertical direction is 

 $work done by horizontal force is W= F\times \ell \sin 45°$
           $Work done by horizontal force =\frac{F\ell }{\sqrt{2}}$---------(1)
          $$\begin{gathered} gain in PE of pendulum = Mgh \\ PE=Mg(\ell -\ell \cos 45°) \\ PE=Mg\ell (1-\frac{1}{\sqrt{2}}) \end{gathered}$$
             $PE =Mg\ell \left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$-----------(2)
        $$\begin{gathered} substitute equation \left(1\right) in \left(2\right) \\ \therefore \frac{F\ell }{\sqrt{2}}=Mg\ell \left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \end{gathered}$$
        $\Rightarrow F=Mg(\sqrt{2}-1)$