Three solid spheres each of mass m and radius R are released from the position shown in figure. The speed of any one sphere at the time of collision would be

$\sqrt{Gm\left(\frac{1}{R}-\frac{2}{d}\right)}$

From conservation of mechanical energy 

$3\left\{\frac{1}{2} m v^2\right\} = 3\left\{\frac{G m^2}{2 R}-\frac{G m^2}{d}\right\}$

$v^2=G m\left\{\frac{1}{R}-\frac{2}{d}\right\}$

∴ $v=\sqrt{G m\left\{\frac{1}{R}-\frac{2}{d}\right\}}$

Hence, (D) is correct.