I've not come across a straight forward method to do so.

What I do is, write the addresses in the binary form.

Now start matching the addresses from left upto the point where all the given addresses agree to be equal.

Count the nber of bits of the matching, which will be the new mask.

In their case,

(I'm expanding just the last two octets, as 192.24 is common to all)

.00000000.00000000/21

.00010000.00000000/20

.00001000.00000000/22

We can clearly see that they are matching upto .000

After counting, 8+8(of 192.24) + 3= 19.

Which would be new mask. And address would be 192.24.0.0/19

What I do is, write the addresses in the binary form.

Now start matching the addresses from left upto the point where all the given addresses agree to be equal.

Count the nber of bits of the matching, which will be the new mask.

In their case,

(I'm expanding just the last two octets, as 192.24 is common to all)

.00000000.00000000/21

.00010000.00000000/20

.00001000.00000000/22

We can clearly see that they are matching upto .000

After counting, 8+8(of 192.24) + 3= 19.

Which would be new mask. And address would be 192.24.0.0/19