Refer to the exhibit.

The correct answer is B .

The study guide explicitly states that slabs are used by the kernel : “The kernel memory slabs are collections of objects with a common purpose. The kernel uses them to store information in memory.”

It also gives the exact calculation method: “Total slab size = available objects x object size” and explains that in the diagnose hardware sysinfo slab output, the columns are active objects , available objects , and object size

From the exhibit:

tcp_session 3 5 1500 ...

available objects = 5

object size = 1500

So:

Total slab size = 5 × 1500 = 7500

That matches option B .

Why the other options are wrong:

A : ip_session 10 10 1408 ... gives 10 × 1408 = 14080 , but slabs are associated with the kernel , not user space

C : ip6_session 5 0 1472 ... gives 0 × 1472 = 0 , not 1472

D : UDPv6 15 10 1408 ... gives 10 × 1408 = 14080 , but again slabs are associated with the kernel , not user space

So the verified answer is B .