Solution

Given block is:

$130.56.0.0/16$

Number of subnets required:

$1024 = 2^{10}$

So, new prefix is:

$/16 + 10 = /26$

For $/26$, each subnet has:

$2^{32-26} = 2^6 = 64$ addresses

So, subnet block size is:

$64$

In the last octet, subnets will be:

$0, 64, 128, 192$

The last subnet of the whole block $130.56.0.0/16$ will be in the last third octet:

$255$

So, last subnet starts at:

$130.56.255.192$

Since block size is $64$, it ends at:

$130.56.255.255$