If R is the radius and H is the height

mass of water in first tube = m = volume × density = \pi R^2 Hρ

                                        surface tension = T = \frac{HρgR}{2}  ................................. (1)

                    Let H' is the height to which water rises in second tube & R' is the radius.

                              As R' = 2R ----------- given

                              Hence mass of water in 2^nd tube,

                                   M' =   \pi R'^{2} H'ρ

                    and

                                        surface Tension = \frac{H'ρgR}{2}............................................(2)

                    surface tension will remain same hence from (1) & (2)

                                           \frac{HρgR}{2} = \frac{H'ρgR}{2}

                                            ∴ HR = H'R'

                                            ∴ HR = H' × 2R

                                           ∴ H = 2H' & H' = (H/2)

                    ∴ mass of water in 2^nd tube = M' = \pi R'^{2} H'ρ

                                                                      = \pi * (2R)^2 * \frac{H}{2} * ρ

                                                                      = 2 \pi R^2Hρ

                                                                 M'  = 2M   

Capillary action:

If you stick the open end of a capillary tube into water, the water will draw itself right up into the tube.

Mass of the water in the capillary tube M = \pi R^2 H ρ

                                  Where R = radius

                                              H = height  

                                              ρ  = Density of liquid    

From this we can write  M \propto R (since  ρ, \pi , and H assume as a constants ).

            If R increases M also increases.