From work energy theorem maximum extension of spring is x_m 

        ${\mathrm{mgx}}_{\mathrm{m}}\sin \mathrm{\theta }=\frac{1}{2}{\mathrm{kx}}_{\mathrm{m}}^2$

or    ${\mathrm{x}}_{\mathrm{m}}=\frac{2\mathrm{mgsin}\mathrm{\theta }}{\mathrm{k}}$

At lower extreme position
        kx_m – mgsinθ – f = ma         ....(i)            
         a = Rα                                ....(ii)

        $\mathrm{fxR}=\frac{{\mathrm{mR}}^2}{2}\mathrm{\alpha }$                  ....(iii)

Thus     $\mathrm{F}=\frac{\mathrm{mgsin}\mathrm{\theta }}{3}=\frac{1\times 10\times 3/5}{3}=2\mathrm{N}$